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TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define int long long\nusing namespace std;\nusing ll = long long;\nconst int MAXN = 200050;\nstruct dktree{int val,l,r,lazy,sum,anscnt;}t[MAXN];\nint ans = 0;\nint nowpos = 1;\nbool dfs(int u)\n{\n if(t[u].anscnt)return true;\n ans++;\n t[u].anscnt++;\n int l = t[u].l,r= t[u].r;\n t[0].sum=0;t[0].lazy=0;\n if(t[l].sum >= t[u].val){if(dfs(l)){ans += t[l].anscnt;};t[u].anscnt += t[l].anscnt;}\n t[r].sum += t[u].lazy;\n t[r].lazy += t[u].lazy;\n t[u].lazy = 0;\n t[0].sum=0;t[0].lazy=0;\n if(t[r].sum >= t[u].val){dfs(r);t[u].anscnt += t[r].anscnt;}\n return false;\n}\nint solve()\n{\n ans = 0;\n nowpos = 1;\n int n,x;cin>>n>>x;\n for(int i = 1 ; i <= n;i++)\n {\n cin>>t[i].val;\n t[i].l=t[i].r=t[i].lazy=t[i].sum=t[i].anscnt = 0;\n }\n t[0].val = 0x3f3f3f3f;\n t[0].l = t[0].r = t[0].lazy = t[0].sum=t[0].anscnt =0;\n t[0].anscnt = 0;\n vector<int>v;\n v.push_back(0);\n for(int i = 1 ; i <= n;i++)\n { //ๅ ๅ
ฅๆฐ่็น\n int tep = 0;\n while(t[i].val >= t[v.back()].val){//lazyๆ ่ฎฐไธ็งป\n tep = v.back();v.pop_back();\n t[tep].sum -= t[tep].lazy;\n t[v.back()].lazy += t[tep].lazy;\n t[0].sum=0;\n t[0].val=0x3f3f3f3f;\n t[0].anscnt = 0;\n t[tep].lazy=0;\n }\n if(v.size()>=2)\n {\n t[v.back()].sum -= t[v.back()].lazy;\n t[v.size()-2].lazy += t[v.back()].lazy;\n }\n t[i].l = tep;\n ans -= t[tep].anscnt;\n t[i].sum = t[tep].sum;\n t[i].lazy = -t[i].val;\n t[v.back()].r = i;\n \n v.push_back(i);\n nowpos = min(nowpos,(ll)v.size()-1);\n for(;nowpos<v.size();nowpos++)\n if(t[v[nowpos]].anscnt == 0)\n break;\n if(nowpos != 1)nowpos--;//ๆๅไธไธชๆซๆ่ฟ็ไฝ็ฝฎ\n t[v[nowpos]].sum += t[i].val;\n t[v[nowpos]].lazy += t[i].val;\n if(nowpos != 1)\n {\n t[v[nowpos-1]].lazy -= t[i].val;\n t[1].sum += t[i].val;\n t[1].lazy += t[i].val;\n }\n ans -= t[v[nowpos]].anscnt;\n t[v[nowpos]].anscnt = 0;\n dfs(v[nowpos]);\n if(i>=x)\n cout<<ans<<\" \";\n }\n cout<<endl;\n return 0;\n}\nsigned main()\n{\n int T;\n T=1;\n cin>>T;\n while(T--)solve();\n return 0;\n}\n/*\n \n \n*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 276036970,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\n#ifdef DEBUG\n#include \"E:/OI/normal/templates/debug.h\"\n#else\n#define dbg(...) (void)0\n#define msg(...) (void)0\n#endif\n#define ll long long\n#define endl '\\n'\n#define PB emplace_back\n#define PPB pop_back\n#define MP make_pair\n#define ALL(Name) Name.begin(),Name.end()\n#define PII pair<int,int>\n#define VI vector<int>\n#define GI greater<int>\n#define fi first\n#define se second\n \nconst int N=200005;\nint n,djfkafkjadsk;\nint fa[N],val[N];\nll pre[N],a[N];\nint find(int x){return fa[x]==x?x:fa[x]=find(fa[x]);}\n \nint main()\n{\n\tios::sync_with_stdio(false),cin.tie(nullptr);\n\tint _;cin>>_;while(_--)\n{\n\tcin>>n>>djfkafkjadsk;\n\tVI stk1,stk2;\n\tstk1.PB(0),a[0]=LLONG_MAX;\n\tfor(int i=1;i<=n;i++)\n\t{\n\t\tcin>>a[i];pre[i]=pre[i-1]+a[i];\n\t\tint lst=0;\n\t\twhile(a[stk1.back()]<=a[i])lst=stk1.back(),stk1.PPB();\n\t\twhile(!stk2.empty()&&stk2.back()>=stk1.size())stk2.PPB();\n\t\tfa[i]=i,val[i]=1+(pre[i-1]-pre[stk1.back()]>=a[i]?val[lst]:0),stk1.PB(i);\n\t\tstk2.PB(stk1.size()-1);\n\t\tVI rem;\n\t\tfor(int l=stk2.size()-1;~l;l--)\n\t\t{\n\t\t\tint j=stk2[l],k=find(stk1[j-1]);\n\t\t\tif(pre[i]-pre[stk1[j-1]]>=a[stk1[j-1]])\n\t\t\t\tfa[stk1[j]]=k,val[k]+=val[stk1[j]];\n\t\t\telse rem.PB(j);\n\t\t}\n\t\tcout<<val[stk1[1]]<<\" \\n\"[i==n];\n\t\treverse(ALL(rem)),stk2=rem;\n\t}\n}\n \n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275709225,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n#define ll long long\n#define pb(x) push_back(x)\n#define mp(x,y) make_pair(x,y)\n#define all(x) x.begin(),x.end()\n#define vf first \n#define vs second\nconst int mod = 1000000007;\n \nmt19937_64 RNG(chrono::steady_clock::now().time_since_epoch().count());\n \nstruct SegMX{\n\tint siz = 1;\n\tvector<pair<int,int>> sums;\n\tvoid init(int n){\n\t\twhile(siz < n)siz*=2;\n\t\t\n\t\tsums.assign(siz * 2, mp(-1,-1));\n\t}\n \n\tpair<int,int> merge(pair<int,int> a , pair<int,int> b){\n\t\tif(a.vf > b.vf)return a;\n\t\telse return b;\n\t}\n \n\tvoid build(vector<int>& a , int x , int lx , int rx){\n\t\tif(rx - lx == 1){\n\t\t\tif(lx < (int)a.size()){\n\t\t\t\tsums[x] = mp(a[lx] ,lx);\n\t\t\t}\n\t\t\treturn;\n\t\t}\n\t\tint m = (lx + rx)/2;\n\t\tbuild(a,2*x+1,lx,m);\n\t\tbuild(a,2*x+2,m,rx);\n\t\tsums[x] = merge(sums[2*x+1],sums[2*x+2]);\n\t}\n \n\tvoid build(vector<int>& a){\n\t\tbuild(a,0,0,siz);\n\t}\n\t\n\tpair<int,int> range_mx(int l , int r , int x , int lx , int rx){\n\t\tif(lx >= l && rx <= r){\n\t\t\treturn sums[x];\n\t\t}\n\t\tif(lx >= r || rx <= l)return mp(-1,-1);\n\t\tint m = (lx + rx) / 2;\n\t\treturn merge(range_mx(l , r , 2 * x + 1 , lx , m) , range_mx(l , r , 2 * x + 2 , m , rx));\n\t}\n\t\n\tpair<int,int> range_mx(int l , int r){\n\t\treturn range_mx(l , r , 0 , 0 , siz);\n\t}\n\t\n};\n \nconst int N = 200004;\nint ans;\nll pre[N];\nSegMX st;\nll cnt[N];\n \nvoid dnc(int l , int r , int mx, int last , int first){\n\tif(r < l)return;\n\tif(pre[r] - (l == 0 ? 0 : pre[l-1]) < mx){\n\t\tif(l == 0){\n\t\t\tlast = r;\n\t\t\tmx = 0;\n\t\t\tfirst = 0;\n\t\t}\n\t\telse return;\n\t}\n\t//cout << l << ' ' << r << ' ' << mx << ' ' << last << '\\n';\n\tpair<int,int> pos = st.range_mx(l , r + 1);\n\tcnt[last]++;\n\tint lb = pos.vs , rb = r;\n\tint fi = pos.vs;\n\twhile(lb <= rb){\n\t\tint mid = (lb + rb)/2;\n\t\tll sum = pre[mid] - (l == 0 ? 0 : pre[l-1]);\n\t\tif(sum >= mx){\n\t\t\tfi = mid;\n\t\t\trb = mid - 1;\n\t\t}\n\t\telse lb = mid + 1;\n\t}\n\tint first1 = max(first , fi);\n\tif(l == 0)first1 = fi;\n\t//if(l == 1)\n //cout << l << ' ' << r << ' ' << fi << '\\n';\n\tif(first1 != 0)cnt[first1-1]--;\n\tdnc(l,pos.vs-1,pos.vf , last , fi);\n\tdnc(pos.vs + 1 , r , pos.vf , last , max(first,fi));\n}\n \nvoid solve(){\n\tint n,x;\n\tcin >> n >> x;\n\tvector<int> arr(n);\n\tfor(int i = 0; i < n; i++)\n\t\tcin >> arr[i];\n\tpre[0] = arr[0];\n\tfor(int i = 1; i < n; i++)pre[i] = pre[i-1] + arr[i];\n\tst.init(n + 1);\n\tst.build(arr);\n\tpair<int,int> pos = st.range_mx(0,n);\n\tfor(int i = 0; i <= n + 1; i++)cnt[i] = 0;\n\tcnt[n-1]++;\n\tif(pos.vs != 0)cnt[pos.vs-1]--;\n\tdnc(0,pos.vs-1 , pos.vf , n-1 , 0);\n\tdnc(pos.vs + 1 , n-1 , pos.vf , n-1 , 0);\n\tfor(int i = n - 1; i >= x-1; i--){\n\t\tcnt[i] += cnt[i + 1];\n\t}\n\tfor(int i = x - 1; i <= n - 1; i++){\n\t\tcout << cnt[i] << ' ';\n\t}\n\tcout << '\\n';\n}\n \nint main(){\n\tios_base::sync_with_stdio(false);\n\tcin.tie(NULL);\n\t\n\tint t;cin >> t;\n\twhile(t--)\n\t\tsolve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275626684,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\nimport math\nimport heapq\nimport itertools\nimport bisect\nimport random\nimport time\nfrom collections import deque\nimport sys\nfrom cmath import exp,pi\nfrom functools import cmp_to_key\ninput=sys.stdin.readline\n \n \ndef update(p,x1,x2):\n idx=p+tree_size\n tree1[idx],tree2[idx]=x1,x2\n while idx>1:\n idx//=2\n tree1[idx]=max(tree1[idx*2],tree1[idx*2+1])\n tree2[idx]=min(tree2[idx*2],tree2[idx*2+1])\n \ndef query1(l,r):\n l+=tree_size\n r+=tree_size\n res=-1\n while l<=r:\n if l&1==1:\n res=max(res,tree1[l])\n l+=1\n if r&1==0:\n res=max(res,tree1[r])\n r-=1\n l//=2\n r//=2\n return res\n \ndef query2(l,r,max_value):\n l+=tree_size\n r+=tree_size\n res=max_value\n while l<=r:\n if l&1==1:\n res=min(res,tree2[l])\n l+=1\n if r&1==0:\n res=min(res,tree2[r])\n r-=1\n l//=2\n r//=2\n return res\n \nt=int(input())\nfor _ in range(t):\n n,x=map(int,input().split())\n arr=list(map(int,input().split()))\n num=[[arr[i],i] for i in range(n)]\n num=sorted(num,key=lambda x:-x[0])\n \n tree_size=1\n while tree_size<n:\n tree_size*=2\n tree1=[-1]*(tree_size*2) #max\n tree2=[10**9]*(tree_size*2) #min\n \n ans=[0]*n\n s=[0]\n for i in arr:\n s.append(s[-1]+i)\n \n ans=[1]\n possible=[0]*n\n possible[0]=1\n update(0,0,0)\n destroyed=-1\n cnt=1\n queue=[]\n \n for i in range(1,n):\n if s[i]<arr[i]:\n for j in range(destroyed+1,i):\n update(0,-1,10**9)\n if possible[j]==1:\n possible[j]=0\n cnt-=1\n possible[i]=1\n update(i,i,i)\n cnt+=1\n destroyed=i-1\n else:\n prev=query1(0,n)\n heapq.heappush(queue,(-arr[i],i))\n while queue:\n v,idx=heapq.heappop(queue)\n #print(i,v,idx)\n if prev+1<=idx<=i:\n l=query1(0,idx-1)\n r=query2(idx+1,n,i+1)\n #print(l,r)\n flag=0\n if l!=-1 and arr[l]<=s[r]-s[l+1]:\n flag=1\n if r!=i+1 and arr[r]<=s[r]-s[l+1]:\n flag=1\n if flag==1:\n possible[idx]=1\n cnt+=1\n update(idx,idx,idx)\n else:\n heapq.heappush(queue,(v,idx))\n break\n \n ans.append(cnt)\n #Qprint(possible)\n print(*ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275610737,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\nimport math\nimport heapq\nimport itertools\nimport bisect\nimport random\nimport time\nfrom collections import deque\nimport sys\nfrom cmath import exp,pi\nfrom functools import cmp_to_key\ninput=sys.stdin.readline\n \n \ndef update(p,x1,x2):\n idx=p+tree_size\n tree1[idx],tree2[idx]=x1,x2\n while idx>1:\n idx//=2\n tree1[idx]=max(tree1[idx*2],tree1[idx*2+1])\n tree2[idx]=min(tree2[idx*2],tree2[idx*2+1])\n \ndef update2(p,x,k):\n idx=p+tree_size\n tree3[idx]=[x,k]\n while idx>1:\n idx//=2\n if tree3[idx*2][0]>tree3[idx*2][1]:\n tree3[idx]=tree3[idx*2]\n else:\n tree3[idx]=tree3[idx*2+1]\n \ndef query1(l,r):\n l+=tree_size\n r+=tree_size\n res=-1\n while l<=r:\n if l&1==1:\n res=max(res,tree1[l])\n l+=1\n if r&1==0:\n res=max(res,tree1[r])\n r-=1\n l//=2\n r//=2\n return res\n \ndef query2(l,r,max_value):\n l+=tree_size\n r+=tree_size\n res=max_value\n while l<=r:\n if l&1==1:\n res=min(res,tree2[l])\n l+=1\n if r&1==0:\n res=min(res,tree2[r])\n r-=1\n l//=2\n r//=2\n return res\n \ndef query3(l,r):\n l+=tree_size\n r+=tree_size\n m=p=-1\n while l<=r:\n if l&1==1:\n if m<tree3[l][0]:\n m,p=tree3[l][0],tree3[l][1]\n l+=1\n if r&1==0:\n if m<tree3[r][0]:\n m,p=tree3[l][0],tree3[l][1]\n r-=1\n l//=2\n r//=2\n return p\n \n \ndef add(p):\n global cnt\n cnt+=1\n update(p,p,p)\n update2(p,-1,-1)\n \ndef remove(p):\n global cnt\n cnt-=1\n update(p,-1,10**9)\n update2(p,arr[p],p)\n \n \nt=int(input())\nfor _ in range(t):\n n,x=map(int,input().split())\n arr=list(map(int,input().split()))\n num=[[arr[i],i] for i in range(n)]\n num=sorted(num,key=lambda x:-x[0])\n \n tree_size=1\n while tree_size<=n:\n tree_size*=2\n tree1=[-1]*(tree_size*2) #max\n tree2=[10**9]*(tree_size*2) #min\n tree3=[[-1,-1] for _ in range(tree_size*2)]\n for i in range(n):\n update2(i,arr[i],i)\n \n ans=[0]*n\n s=[0]\n for i in arr:\n s.append(s[-1]+i)\n \n ans=[1]\n possible=[0]*n\n possible[0]=1\n update(0,0,0)\n destroyed=-1\n cnt=1\n \n for i in range(1,n):\n if s[i]<arr[i]:\n for j in range(destroyed+1,i):\n if possible[j]==1:\n possible[j]=0\n remove(j)\n possible[i]=1 \n add(i)\n destroyed=i-1\n else:\n prev=query1(0,n)\n queue=deque([(prev,i+1)])\n while queue:\n l,r=queue.popleft()\n if l+1>=r:\n continue\n \n idx=query3(l+1,r-1)\n #print(i,l,r,idx)\n flag=0\n if l!=-1 and arr[l]<=s[r]-s[l+1]:\n flag=1\n if r!=i+1 and arr[r]<=s[r]-s[l+1]:\n flag=1\n if flag==1:\n add(idx)\n possible[idx]=1\n queue.append((l,idx))\n queue.append((idx,r))\n \n #print(possible)\n ans.append(cnt)\n \n print(*ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275626096,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\nimport math\nimport heapq\nimport itertools\nimport bisect\nimport random\nimport time\nfrom collections import deque\nimport sys\nfrom cmath import exp,pi\nfrom functools import cmp_to_key\ninput=sys.stdin.readline\n \n \ndef update(p,x1,x2):\n idx=p+tree_size\n tree1[idx],tree2[idx]=x1,x2\n while idx>1:\n idx//=2\n tree1[idx]=max(tree1[idx*2],tree1[idx*2+1])\n tree2[idx]=min(tree2[idx*2],tree2[idx*2+1])\n \ndef update2(p,x,k):\n idx=p+tree_size\n tree3[idx]=[x,k]\n while idx>1:\n idx//=2\n if tree3[idx*2][0]>tree3[idx*2+1][0]:\n tree3[idx]=tree3[idx*2]\n else:\n tree3[idx]=tree3[idx*2+1]\n \ndef query1(l,r):\n l+=tree_size\n r+=tree_size\n res=-1\n while l<=r:\n if l&1==1:\n res=max(res,tree1[l])\n l+=1\n if r&1==0:\n res=max(res,tree1[r])\n r-=1\n l//=2\n r//=2\n return res\n \ndef query2(l,r,max_value):\n l+=tree_size\n r+=tree_size\n res=max_value\n while l<=r:\n if l&1==1:\n res=min(res,tree2[l])\n l+=1\n if r&1==0:\n res=min(res,tree2[r])\n r-=1\n l//=2\n r//=2\n return res\n \ndef query3(l,r):\n l+=tree_size\n r+=tree_size\n m=p=-1\n while l<=r:\n if l&1==1:\n if m<tree3[l][0]:\n m,p=tree3[l][0],tree3[l][1]\n l+=1\n if r&1==0:\n if m<tree3[r][0]:\n m,p=tree3[l][0],tree3[l][1]\n r-=1\n l//=2\n r//=2\n #print(m,p)\n return p\n \n \ndef add(p):\n global cnt\n cnt+=1\n update(p,p,p)\n update2(p,-1,-1)\n \ndef remove(p):\n global cnt\n cnt-=1\n update(p,-1,10**9)\n update2(p,arr[p],p)\n \n \nt=int(input())\nfor _ in range(t):\n n,x=map(int,input().split())\n arr=list(map(int,input().split()))\n num=[[arr[i],i] for i in range(n)]\n num=sorted(num,key=lambda x:-x[0])\n \n tree_size=1\n while tree_size<=n:\n tree_size*=2\n tree1=[-1]*(tree_size*2) #max\n tree2=[10**9]*(tree_size*2) #min\n tree3=[[-1,-1] for _ in range(tree_size*2)]\n for i in range(n):\n update2(i,arr[i],i)\n \n cnt=0\n ans=[0]*n\n s=[0]\n for i in arr:\n s.append(s[-1]+i)\n \n ans=[1]\n possible=[0]*n\n possible[0]=1\n add(0)\n destroyed=-1\n \n for i in range(1,n):\n if s[i]<arr[i]:\n for j in range(destroyed+1,i):\n if possible[j]==1:\n possible[j]=0\n remove(j)\n possible[i]=1 \n add(i)\n destroyed=i-1\n else:\n prev=query1(0,n)\n queue=deque([(prev,i+1)])\n while queue:\n l,r=queue.popleft()\n if l+1>=r:\n continue\n \n idx=query3(l+1,r-1)\n #print(i,l,r,idx)\n flag=0\n if l!=-1 and arr[l]<=s[r]-s[l+1]:\n flag=1\n if r!=i+1 and arr[r]<=s[r]-s[l+1]:\n flag=1\n if flag==1:\n add(idx)\n possible[idx]=1\n queue.append((l,idx))\n queue.append((idx,r))\n \n #print(possible)\n ans.append(cnt)\n \n print(*ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275629430,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class E {\n\t\n\tstatic IOHandler sc = new IOHandler();\n\tstatic PrintWriter out = new PrintWriter(System.out);\n\t\n\tpublic static void main(String[] args) throws IOException {\n\t\t\n new Thread(null, new Solution(), \"solution\", 1 << 26 ).start();\n }\n\t\n\tprivate static class Solution implements Runnable {\n \n\t\tpublic void run() {\n\t\t\t// TODO Auto-generated method stub\n\t\t\tint testCases = sc.nextInt();\n\t\t\t\n\t\t\tfor (int i = 1; i <= testCases; ++i) {\n\t\t\t\tsolve(i);\n\t\t\t}\n\t\t\t\n\t\t\tout.flush();\n\t\t}\n\t\t\n\t\t\n\t\tprivate void advice() {\n\t\t\t// Read question and Test cases thoroughly\n\t\t\t// Check for overflow , Avoid Integer == \n\t\t\t// Handle invalid cases\n\t\t\t// Think before solving\n\t\t\t// Search for concepts\n\t\t\t// expected alg like seg tree, trie, binSearch, sqrt decom ... \n\t\t\t\n\t\t\t// Source of Error\n\t\t\t// Int overflow, Not returning after special case\n\t\t\t// Not taking mod properly\n\t\t\t// sorting input groups after reading certain values\n\t\t\t// Not using the right vairable eg idx in method for i in loop\n\t\t\t// Accounting for special case in general logic. eg special in dp\n\t\t\t// for BIT. Do not use tree maps with update query\n\t\t\t// Assuming Constraints on an unbound variable\n\t\t\t// Not properly calculating time complexity\n\t\t}\n\t\t\n\t\t\n\t\tprivate void solve(int t) {\n\t\t\tint n = sc.nextInt();\n\t\t\t\n\t\t\tint k = sc.nextInt();\n\t\t\t\n\t\t\tint [] arr = sc.readArray2(n);\n\t\t\t\n\t\t\t\n\t\t\tlong [] prefix = new long [n + 1];\n\t\t\t\n\t\t\tint [] next = new int [n + 1];\n\t\t\t\n\t\t\tlong sum = 0;\n\t\t\t\n\t\t\tfor (int i = 0; i <= n; ++i) {\n\t\t\t\tsum += arr[i];\n\t\t\t\tprefix[i] = sum;\n\t\t\t}\n\t\t\t\n\t\t\tnext[n] = n + 1;\n\t\t\t\n\t\t\tfor (int i = 1; i <= n; ++i) {\n\t\t\t\tint max = n + 1;\n\t\t\t\tint min = i;\n\t\t\t\tint mid;\n\t\t\t\t\n\t\t\t\twhile (max - min > 1) {\n\t\t\t\t\tmid = (max + min) / 2;\n\t\t\t\t\t\n\t\t\t\t\tif (prefix[mid] - prefix[i] >= arr[i]) {\n\t\t\t\t\t\tmax = mid;\n\t\t\t\t\t} else {\n\t\t\t\t\t\tmin = mid;\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t\t\n\t\t\t\tnext[i] = max;\n\t\t\t}\n\t\t\t\n\t\t\tint [] dp = new int [n + 1];\n\t\t\t\n\t\t\tArrays.fill(dp, -1);\n\t\t\t\n\t\t\tsolve(arr, prefix, next, dp, 1, n);\n\t\t\t\n\t\t\tint last = -1;\n\t\t\t\n\t\t\tStringBuilder sb = new StringBuilder();\n\t\t\t\n\t\t\tfor (int i = 1; i <= n; ++i) {\n\t\t\t\tif (dp[i] >= 0)\n\t\t\t\t\tlast = dp[i];\n\t\t\t\t\n\t\t\t\tsb.append(last + \" \");\n\t\t\t}\n\t\t\t\n\t\t\t\n\t\t\tout.println(sb);\n\t\t}\n\t\t\n\t\t\n\t\tprivate int solve(int [] arr, long [] prefix, int [] next,\n\t\t\t\tint [] dp, int l, int r) {\n\t\t\t\n\t\t\tint current = 1;\n\t\t\tint pos = l;\n\t\t\t\n\t\t\t\n\t\t\tint stop = next[l];\n\t\t\t\n\t\t\tint left, right;\n\t\t\tint prev = l;\n\t\t\tint valid;\n\t\t\t\n\t\t\tlong sum;\n\t\t\tlong added;\n\t\t\t\n\t\t\tif (l == 1) {\n\t\t\t\tdp[1] = current;\n\t\t\t}\n\t\t\t\n\t\t\twhile (stop <= r) {\n\t\t\t\tleft = prev + 1;\n\t\t\t\tright = stop;\n\t\t\t\t\n\t\t\t\tsum = prefix[stop - 1] - prefix[l - 1];\n\t\t\t\tadded = arr[stop];\n\t\t\t\t\n\t\t\t\tif (added > sum) {\n\t\t\t\t\tcurrent = 1;\n\t\t\t\t}else {\n\t\t\t\t\tcurrent += solve(arr, prefix, next, dp, left, right);\n\t\t\t\t}\n\t\t\t\t\n\t\t\t\tif (l == 1) {\n\t\t\t\t\tdp[stop] = current;\n\t\t\t\t}\n\t\t\t\t\n\t\t\t\tprev = stop;\n\t\t\t\tstop = next[prev];\n\t\t\t}\n\t\t\t\n\t\t\treturn current;\n\t\t}\n\t\t\n\t\t\n\t\t\n\t}\n\t\n\tprivate static class IOHandler {\n BufferedReader br;\n StringTokenizer st;\n \n public IOHandler() {\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n \n String next() {\n while (st == null || !st.hasMoreElements()) {\n try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n int [] readArray(int n) {\n \t int [] res = new int [n];\n \t \n \t for (int i = 0; i < n; ++i)\n \t\t res[i] = nextInt();\n \t \n \t return res;\n }\n \n int [] readArray2(int n) {\n \t int [] res = new int [n + 1];\n \t \n \t for (int i = 1; i <= n; ++i)\n \t\t res[i] = nextInt();\n \t \n \t return res;\n }\n \n int [][] readGraph(int n, int m, int c) {\n \t int [][] graph = new int [n + 1][];\n \t \n \t int [] count = new int [n + 1];\n \t int [][] arr = new int [m][1 + c];\n \t \n \t int a, b;\n \t \n \t for (int i = 0; i < m; ++i) {\n \t\t a = sc.nextInt();\n \t\t b = sc.nextInt();\n \t\t \n \t\t arr[i][0] = a;\n \t\t arr[i][1] = b;\n \t\t \n \t\t ++count[a];\n \t\t ++count[b];\n \t\t \n \t\t for (int j = 2; j <= c; ++j) {\n \t\t\t arr[i][j] = sc.nextInt();\n \t\t }\n \t }\n \t \n \t for (int i = 0; i <= n; ++i) {\n \t\t graph[i] = new int [count[i] * c];\n \t\t count[i] = 0;\n \t }\n \t int swap;\n \t \n \t for (int [] tArr : arr) {\n \t\t for (int j = 1; j < tArr.length; ++j) {\n \t\t\t graph[tArr[0]][count[tArr[0]]++] = tArr[j]; \n \t\t }\n \t\t \n \t\t swap = tArr[0];\n \t\t tArr[0] = tArr[1];\n \t\t tArr[1] = swap;\n \t\t \n \t\t for (int j = 1; j < tArr.length; ++j) {\n \t\t\t graph[tArr[0]][count[tArr[0]]++] = tArr[j]; \n \t\t }\n \t }\n \t \n \t return graph;\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n String nextLine(){\n String str = \"\";\n\t\t try {\n\t\t str = br.readLine();\n\t\t } catch (IOException e) {\n\t\t e.printStackTrace();\n\t\t }\n\t\t return str;\n }\n \n }\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275666790,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n//\n// Created by codicon on 8/24/2024 at 3:00 PM.\n//\n \n \n#include <bits/stdc++.h>\n \nusing namespace std;\n \n#define ar array\ntypedef long long ll;\ntypedef int uci;\n#define int long long\n#define F first\n#define S second\n#define rep(i, a, b) for(int i = a; i < (b); ++i)\n#define irep(i, a, b) for(int i = a; i > (b); --i)\n#define all(x) begin(x), end(x)\n#define sz(x) (int)(x).size()\ntypedef pair<int, int> pii;\n#define ve vector\ntypedef ve<int> vi;\ntypedef ve<vi> vvi;\n \nvoid solve() {\n int n, x; cin >> n >> x;\n vi a(n+1); rep(i, 1, n+1) cin >> a[i];\n \n vi psum = a; partial_sum(all(psum), psum.begin());\n vi nge(n+1, n+1), pge(n+1, 0);\n \n vi stk;\n rep(i, 1, n+1) {\n while (!empty(stk) && a[i] >= a[stk.back()]) {\n nge[stk.back()] = i;\n stk.pop_back();\n }\n stk.push_back(i);\n }\n stk.clear();\n irep(i, n, 0) {\n while (!empty(stk) && a[i] >= a[stk.back()]) {\n pge[stk.back()] = i;\n stk.pop_back();\n }\n stk.push_back(i);\n }\n \n vvi s_events(n+2), e_events(n+2);\n \n rep(i, 1, n+1) {\n if (i+1 == nge[i]) continue;\n \n int l = i+1, r = nge[i];\n \n while (l != r) {\n int m = (l + r) / 2;\n if (psum[m] - psum[i] >= a[i]) {\n r = m;\n }\n else {\n l = m + 1;\n }\n }\n \n s_events[i+1].push_back(r == nge[i] ? n+1 : r);\n s_events[nge[i]].push_back(-(r == nge[i] ? n+1 : r));\n }\n \n rep(i, 1, n+1) {\n if (pge[i] + 1 == i) continue;\n \n if (psum[i-1] - psum[pge[i]] < a[i]) {\n e_events[pge[i]+1].push_back(i);\n e_events[i].push_back(-i);\n }\n }\n \n vi ans(n+2);\n \n multiset<int, greater<>> s;\n multiset<int> e;\n rep(i, 1, n+1) {\n for (auto j: s_events[i]) {\n if (j < 0) s.extract(-j);\n else s.insert(j);\n }\n for (auto j: e_events[i]) {\n if (j < 0) e.extract(-j);\n else e.insert(j);\n }\n \n int start = empty(s) ? i : *s.begin();\n int end = empty(e) ? n+1 : *e.begin();\n if (start < end) {\n ans[start]++;\n ans[end]--;\n }\n }\n \n partial_sum(all(ans), ans.begin());\n rep(i, x, n+1) {\n cout << ans[i] << ' ';\n }\n cout << '\\n';\n}\n \nuci main() {\n cin.tie(0)->sync_with_stdio(0);\n int t; cin >> t;\n while (t--) solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 277973685,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n#ifdef _DEBUG\n #include \"debug/all.h\"\n using namespace debug;\n#else\n #include \"bits/stdc++.h\"\n #define dbg(...) 0;\n#endif\n \n#define int long long\n#define pii pair<int, int>\n#define f first\n#define s second\n#define all(x) (x).begin(), (x).end()\n \nusing namespace std;\n \nvoid solve(){\n int n, x;\n cin >> n >> x;\n vector<int> a(n);\n vector<int> pref(1);\n for(auto &x: a) {\n cin >> x;\n pref.push_back(x + pref.back());\n }\n auto get_sum = [&](int l, int r) {\n return pref[r + 1] - pref[l];\n };\n // okay\n // let's think\n // when we get the a[i]\n // we start going through the indices in the stack\n // when we get to any j\n // k < j < i\n // so it is sandwiched between two bigger values (or it isn't when all elements left of it are less than a[j])\n // so we choose the smaller\n // if it's i, then we have to remove the smaller value from the answer for now (if it's on) and add the whole subtree to the i\n // otherwise we remove the answer only if k is not on\n // well to be fair the behavior is consistent with the tree being turned on or off\n // also if the j is about to be deleted, it's faith has already been sealed\n // and after we are done with it, ehh, we do what exactly?\n // std::set of values that are still not on\n \n int ans = 0;\n vector<int> stc; // stack\n vector<vector<int>> add(n + 2); // add[i] - all \n vector<int> sub(n, 1), checked(n);\n set<int> off, pending;\n for(int i = 0; i < n; i++) {\n for(auto x: add[i]) {\n auto it = off.find(x);\n if(it != off.end()) {\n off.erase(x);\n pending.insert(x);\n }\n }\n while(!pending.empty() && (off.empty() || *off.begin() > *pending.begin())){\n ans += sub[*pending.begin()];\n checked[*pending.begin()] = 1;\n pending.erase(pending.begin());\n }\n int subtracted_chain = 0;\n while(stc.size() && a[stc.back()] <= a[i]) {\n int j = stc.back();\n int chained = checked[j] * (sub[j] - subtracted_chain);\n ans -= chained;\n subtracted_chain += chained;\n stc.pop_back();\n auto [mn, mn_ind] = stc.empty() ? pii{a[i], i} : min(pii{a[i], i}, pii{a[stc.back()], stc.back()});\n int l = stc.empty() ? 0ll : stc.back() + 1, r = i - 1;\n if(get_sum(l, r) >= mn) {\n sub[mn_ind] += sub[j];\n }\n else {\n subtracted_chain = 0;\n }\n pending.erase(j);\n off.erase(j);\n }\n if(stc.empty()) {\n ans += sub[i];\n checked[i] = 1;\n }\n else {\n int req = a[stc.back()];\n int where = lower_bound(all(pref), req + pref[stc.back() + 1]) - pref.begin() - 1;\n if(where <= i) {\n ans += sub[i];\n checked[i] = 1;\n }\n else {\n add[where].push_back(i);\n }\n }\n stc.push_back(i);\n if(!checked[i]) off.insert(i);\n cout << ans << ' ';\n }\n cout << '\\n';\n}\n \nsigned main() {\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n int t = 1;\n cin >> t;\n while(t--){\n solve();\n }\n}\n// edge cases are crazy\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 280167656,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing ll = long long;\nconst int N = 20;\n \nvoid charming() {\n int n, x; std::cin >> n >> x;\n std::vector<int> a(n);\n std::vector<ll> pre(n);\n for (int i = 0; i < n; ++i) {\n std::cin >> a[i];\n pre[i] = (i - 1 >= 0 ? pre[i - 1] : 0) + a[i];\n }\n \n std::vector<std::array<int, N> > mx(n);\n for (int i = 0; i < n; ++i) mx[i][0] = a[i];\n for (int j = 1; j < N; ++j) {\n for (int i = 0; i < n; ++i) {\n if (i + (1 << j) <= n) {\n mx[i][j] = std::max(mx[i][j - 1], mx[i + (1 << (j - 1))][j - 1]);\n }\n else mx[i][j] = mx[i][j - 1];\n }\n }\n \n auto Expand = [&](int l, int r) -> int {\n /* \n Given a range which means thar all elements whose index is between l and r are merged to one element.\n return the new range.\n */\n \n while (l > 0) {\n int old_l = l;\n \n ll val = pre[r] - (l - 1 >= 0 ? pre[l - 1] : 0);\n if (val >= 1e9) {l = 0; break;}\n \n for (int j = N - 1; j >= 0; --j) {\n if (l - 1 - (1 << j) + 1 >= 0) {\n if (mx[l - (1 << j)][j] <= val) l -= (1 << j);\n }\n }\n \n if (l == old_l) break;\n }\n \n return l;\n };\n \n int ans = 0;\n std::multiset<int> s;\n // ans: How many i can be the winner.\n // s: For all i which can't be the winner, but can merge all element on the right side,\n // maintain its left side.\n \n // std::vector<int> pre(n + 2), nxt(n + 2);\n // int head = n, tail = n + 1;\n \n for (int i = 0; i < n; ++i) {\n if (i > 0 && pre[i - 1] < a[i]) {\n ans = 0;\n s.clear();\n }\n std::vector<int> vec;\n while (i > 0 && !s.empty()) {\n int old_l = *s.begin();\n s.erase(s.find(old_l));\n assert(old_l > 0);\n ll val = pre[i - 1] - pre[old_l - 1];\n if (val < a[i]) break;\n int new_l = Expand(old_l, i);\n if (new_l == 0) ++ans;\n else {\n assert(new_l <= old_l);\n vec.emplace_back(new_l);\n }\n }\n for (int x : vec) s.insert(x);\n \n int l = Expand(i, i);\n if (l == 0) ++ans;\n else s.insert(l);\n \n // std::cout << i << \" \" << ans << \" \" << (int)s.size() << ' ';\n // for (int x : s) std::cout << x << ' ';\n // std::cout << '\\n';\n \n std::cout << ans << \" \\n\"[i == n - 1];\n }\n}\n \nsigned main() {\n std::ios_base::sync_with_stdio(false);\n std::cin.tie(NULL);\n std::cout.tie(NULL);\n int t; std::cin >> t;\n while (t--) charming();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275691689,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\nimport math\nimport heapq\nimport itertools\nimport bisect\nimport random\nimport time\nfrom collections import deque\nimport sys\nfrom cmath import exp,pi\nfrom functools import cmp_to_key\ninput=sys.stdin.readline\n \n \ndef update(p,x1,x2):\n idx=p+tree_size\n tree1[idx],tree2[idx]=x1,x2\n while idx>1:\n idx//=2\n tree1[idx]=max(tree1[idx*2],tree1[idx*2+1])\n tree2[idx]=min(tree2[idx*2],tree2[idx*2+1])\n \ndef query1(l,r):\n l+=tree_size\n r+=tree_size\n res=-1\n while l<=r:\n if l&1==1:\n res=max(res,tree1[l])\n l+=1\n if r&1==0:\n res=max(res,tree1[r])\n r-=1\n l//=2\n r//=2\n return res\n \ndef query2(l,r,max_value):\n l+=tree_size\n r+=tree_size\n res=max_value\n while l<=r:\n if l&1==1:\n res=min(res,tree2[l])\n l+=1\n if r&1==0:\n res=min(res,tree2[r])\n r-=1\n l//=2\n r//=2\n return res\n \nt=int(input())\nfor _ in range(t):\n n,x=map(int,input().split())\n arr=list(map(int,input().split()))\n num=[[arr[i],i] for i in range(n)]\n num=sorted(num,key=lambda x:-x[0])\n \n tree_size=1\n while tree_size<=n:\n tree_size*=2\n tree1=[-1]*(tree_size*2) #max\n tree2=[10**9]*(tree_size*2) #min\n \n ans=[0]*n\n s=[0]\n for i in arr:\n s.append(s[-1]+i)\n \n ans=[1]\n possible=[0]*n\n possible[0]=1\n update(0,0,0)\n destroyed=-1\n cnt=1\n queue=[]\n \n for i in range(1,n):\n if s[i]<arr[i]:\n for j in range(destroyed+1,i):\n update(0,-1,10**9)\n if possible[j]==1:\n possible[j]=0\n cnt-=1\n possible[i]=1\n update(i,i,i)\n cnt+=1\n destroyed=i-1\n else:\n prev=query1(0,n)\n heapq.heappush(queue,(-arr[i],i))\n while queue:\n v,idx=heapq.heappop(queue)\n #print(i,v,idx)\n if prev+1<=idx<=i:\n l=query1(0,idx-1)\n r=query2(idx+1,n,i+1)\n #print(l,r)\n flag=0\n if l!=-1 and arr[l]<=s[r]-s[l+1]:\n flag=1\n if r!=i+1 and arr[r]<=s[r]-s[l+1]:\n flag=1\n if flag==1:\n possible[idx]=1\n cnt+=1\n update(idx,idx,idx)\n else:\n heapq.heappush(queue,(v,idx))\n break\n \n ans.append(cnt)\n #Qprint(possible)\n print(*ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275611500,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define ll int\n#define LL long long\n#define ls p<<1\n#define rs p<<1|1\n#define Ma 1000005\n#define mod 1000000007\n#define PLL pair<ll,ll>\n#define PDD pair<double,double>\n#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);\n#define fi first\n#define se second\n#define N 61\n#define pb push_back\n#define ld long double\n#define all(x) x.begin(),x.end()\n#define inf 1e18\n \nusing namespace std;\nll n,m,k;\nLL sum[Ma];\nll a[Ma];\n \n \nLL ask(ll l,ll r)\n{\n\treturn sum[r]-sum[l-1];\n}\n \n \nvoid sol()\n{\n\tcin>>n>>m;\n\tfor (ll i=1;i<=n;i++)\n\t\tcin>>a[i],sum[i]=sum[i-1]+a[i];\n\tset <PLL> mp;\n\tmp.insert({n,1});\n\tprintf(\"1 \");\n\tfor (ll i=2;i<=n;i++)\n\t{\n\t\tset <PLL> mp2=mp;\n\t//\tprintf(\"i=%lld\\n\\n\",i);\n\t\tfor(auto z:mp)\n\t\t{\n\t\t\tif (ask(n-z.fi+1,i-1)<a[i])\n\t\t\t{\n\t\t\t//\tprintf(\"GO erase z=%lld %lld\\n\",z.fi,z.se);\n\t\t\t\tmp2.erase(mp2.find(z));\n\t\t\t\t//printf(\"OK\\n\");\n\t\t\t\tcontinue;\n\t\t\t}\n\t\t\tif (z.fi==n)\n\t\t\t\tcontinue;\n\t\t\t//printf(\"OK z=%lld %lld\\n\",n-z.fi+1,z.se);\n\t\t\tif (ask(n-z.fi+1,i)>=a[n-z.fi])\n\t\t\t{\n\t\t\t\tmp2.erase(mp2.find(z));\n\t\t\t\tPLL ne=(*mp.upper_bound(z));\n\t\t\t//\tprintf(\"UP z=%lld %lld\\n\",n-ne.fi+1,ne.se);\n\t\t\t\tmp2.erase(mp2.find(ne));\n\t\t\t\t//printf(\"OK\\n\");\n\t\t\t\tmp2.insert({ne.fi,ne.se+z.se});\n\t\t\t\t//printf(\"OK2\\n\");\n\t\t\t}\n\t\t}\n\t\t//printf(\"OK\\n\");\n\t\tPLL add={n-i+1,1};\n\t\tif (ask(n-add.fi+1,i)<a[i-1])\n\t\t{\n\t\t\tmp2.insert(add);\n\t\t}\n\t\telse\n\t\t{\n\t\t\tadd={(*mp.begin()).fi,1};\n\t\t\tfor (auto z:mp)\n\t\t\t{\n\t\t\t\t\n\t\t\t\tif (ask(n-add.fi+1,i)<a[n-add.fi])\n\t\t\t\t{\n\t\t\t\t\tauto pp=(mp2.lower_bound({add.fi,0}));\n\t\t\t\t\tPLL p=*pp;\n\t\t\t\t\tif (pp!=mp2.end()&&p.fi==add.fi)\n\t\t\t\t\t{\n\t\t\t\t\t\tmp2.erase(mp2.find(p));\n\t\t\t\t\t\tmp2.insert({p.fi,p.se+1});\n\t\t\t\t\t}\n\t\t\t\t\telse\n\t\t\t\t\t\tmp2.insert(add);\n\t\t\t\t\tbreak;\t\n\t\t\t\t}\n\t\t\t\telse\n\t\t\t\t\tadd.fi=z.fi;\n\t\t\t}\n\t\t\tif (add.fi==n)\n\t\t\t{\n\t\t\t\tif (mp2.empty())\n\t\t\t\t\tmp2.insert({n,1});\n\t\t\t\telse\n\t\t\t\t{\n\t\t\t\t\tauto p=*(prev(mp2.end()));\n\t\t\t\t\tmp2.erase(mp2.find(p));\n\t\t\t\t\tmp2.insert({p.fi,p.se+1});\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t\t\t\n\t\tswap(mp,mp2);\t\t\n\t\tPLL z=*prev(mp.end());\n\t\t/*for (auto z:mp)\n\t\t\tprintf(\"%lld %lld\\n\",z.fi,z.se);\n\t\tprintf(\"w=%d\\n\",z.se);*/\n\t\tif (mp.size()>40)\n\t\t{\n\t\t\tprintf(\"-1\\n\");\n\t\t\treturn;\n\t\t}\n\t\tprintf(\"%d \",z.se);\n\t}\n\tprintf(\"\\n\");\n}\n \nint main()\n{\n\tIOS\n\tll tt=1;\n\tcin>>tt;\n\twhile (tt--)\n\t\tsol();\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275643489,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n// cod prost cod prost\n// cod prost cod prost\n// cod prost cod prost\n// cod prost cod prost\n#include <bits/stdc++.h>\nusing namespace std;\n \nusing ll = long long;\n#define int ll\n \n#define endl '\\n'\n#define pb push_back\nusing pi = array<int, 2>;\n \nstruct DSU {\n int n;\n vector<int> par, size;\n stack<pi> stk;\n \n DSU(int n) {\n this->n = n;\n par.assign(n + 1, 0);\n size.assign(n + 1, 1);\n for (int i = 1; i <= n; ++i) {\n par[i] = i;\n }\n }\n \n int find(int a) {\n while (a != par[a]) {\n a = par[a];\n }\n return a;\n }\n void unite(int a, int b) {\n //cerr << \"UNITE \" << a << \" \" << b << endl;\n a = find(a);\n b = find(b);\n if (a == b) {\n //cerr << \"BAD\\n\";\n stk.push({-1, -1});\n return;\n }\n \n if (size[a] < size[b]) {\n swap(a, b);\n }\n par[b] = a;\n size[a] += size[b];\n stk.push({a, b});\n }\n void rollback() {\n //cerr << \"ROLLBACK\\n\";\n if (stk.empty()) {\n //cerr << \"!!BAD\\n\";\n return;\n }\n \n auto [to, from] = stk.top();\n stk.pop();\n if (to == -1) {\n return;\n }\n size[to] -= size[from];\n par[from] = from;\n }\n};\n \nvoid solve() {\n int n, X;\n cin >> n >> X;\n \n vector<int> a(n + 1);\n vector<ll> pref(n + 1);\n for (int i = 1; i <= n; ++i) {\n cin >> a[i];\n pref[i] = pref[i - 1] + a[i];\n }\n \n vector<int> pref_max(n + 1, 1);\n for (int i = 1; i <= n; ++i) {\n pref_max[i] = pref_max[i - 1];\n if (a[i] > a[pref_max[i]]) {\n pref_max[i] = i;\n }\n }\n \n vector<int> left_geq(n + 1), right_gt(n + 1);\n stack<int> stk;\n for (int i = 1; i <= n; ++i) {\n while (!stk.empty() && a[stk.top()] < a[i]) {\n stk.pop();\n }\n if (!stk.empty()) {\n left_geq[i] = stk.top();\n }\n stk.push(i);\n }\n while (!stk.empty()) {\n stk.pop();\n }\n for (int i = n; i >= 1; --i) {\n while (!stk.empty() && a[stk.top()] <= a[i]) {\n stk.pop();\n }\n if (!stk.empty()) {\n right_gt[i] = stk.top();\n }\n stk.push(i);\n }\n \n vector<int> parent(n + 1);\n vector<vector<int>> adj(n + 1);\n for (int i = 1; i <= n; ++i) {\n parent[i] = left_geq[i];\n if (parent[i] == 0 || (right_gt[i] != 0 && a[right_gt[i]] < a[left_geq[i]])) {\n parent[i] = right_gt[i];\n }\n \n adj[parent[i]].pb(i);\n }\n \n vector<int> earliest(n + 1, n + 1);\n auto dfs1 = [&](auto &&self, int u) -> void {\n ll S = pref[u] - pref[left_geq[u]];\n int st = u, dr = right_gt[u] - 1;\n if (right_gt[u] == 0) {\n dr = n;\n }\n \n while (st <= dr) {\n int mid = (st + dr) / 2;\n ll new_S = S + pref[mid] - pref[u];\n \n if (new_S >= a[parent[u]]) {\n earliest[u] = mid;\n dr = mid - 1;\n } else {\n st = mid + 1;\n }\n }\n \n for (int v : adj[u]) {\n self(self, v);\n }\n };\n dfs1(dfs1, 0);\n \n vector<vector<int>> here(4 * (n + 1));\n auto add = [&](auto &&self, int u, int st, int dr, int l, int r, int i) -> void {\n if (l > r) {\n return;\n }\n if (st >= l && dr <= r) {\n here[u].pb(i);\n return;\n }\n \n int mid = (st + dr) / 2;\n if (l <= mid) {\n self(self, 2 * u, st, mid, l, r, i);\n }\n if (mid < r) {\n self(self, 2 * u + 1, mid + 1, dr, l, r, i);\n }\n };\n \n for (int i = 1; i <= n; ++i) {\n // o sa am [i; p[i]) si [earliest[i]; n]\n //cout << i << \" | \" << parent[i] << \" | \" << earliest[i] << endl;\n add(add, 1, 1, n, i, parent[i] - 1, i);\n add(add, 1, 1, n, earliest[i], n, i);\n }\n \n vector<int> ans(n + 1), active(n + 1);\n DSU dsu(n);\n auto dfs2 = [&](auto &&self, int u, int st, int dr) -> void {\n //cerr << \"ENTER \" << st << \" \" << dr << endl;\n \n for (int i : here[u]) {\n ++active[i];\n if (active[i] != 1) {\n continue;\n }\n \n if (active[parent[i]]) {\n dsu.unite(i, parent[i]);\n }\n for (int v : adj[i]) {\n if (active[v]) {\n dsu.unite(i, v);\n }\n }\n }\n \n if (st == dr) {\n int r = pref_max[st];\n ans[st] = dsu.size[dsu.find(r)];\n } else {\n int mid = (st + dr) / 2;\n self(self, 2 * u, st, mid);\n self(self, 2 * u + 1, mid + 1, dr);\n }\n \n for (int i : here[u]) {\n --active[i];\n if (active[i] != 0) {\n continue;\n }\n \n if (active[parent[i]]) {\n dsu.rollback();\n }\n for (int v : adj[i]) {\n if (active[v]) {\n dsu.rollback();\n }\n }\n }\n \n //cerr << \"EXIT \" << st << \" \" << dr << endl;\n };\n dfs2(dfs2, 1, 1, n);\n \n for (int i = 1; i <= n; ++i) {\n cout << ans[i] << \" \";\n }\n}\n \nint32_t main() {\n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n \n int T;\n cin >> T;\n \n while (T--) {\n solve();\n cout << endl;\n }\n}\n \n/*\n \n*/\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275628431,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\n \nusing namespace std;\nusing namespace __gnu_pbds;\nusing ll = long long;\ntemplate<class T> using orderedset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;\nint dx[] = {0, 0, 1, -1, 1, 1, -1, -1};\nint dy[] = {1, -1, 0, 0, 1, -1, 1, -1};\n \nvoid File() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr), cout.tie(nullptr);\n#ifndef ONLINE_JUDGE\n freopen(\"input.txt\", \"r\", stdin);\n freopen(\"output.txt\", \"w\", stdout);\n freopen(\"errors.txt\", \"w\", stderr);\n#else\n \n#endif\n}\n \nconst int MOD = 998244353;\n \ntemplate<typename T, class FUN>\nstruct sparseTable {\n vector<vector<long long>> sp;\n vector<int> _lg;\n FUN calc;\n \n explicit sparseTable(vector<T> &a, FUN calc) : calc(calc) {\n int n = (int) a.size(), max_log = 32 - __builtin_clz(n);\n sp = vector<vector<long long>>(n, vector<long long>(max_log));\n _lg.assign(n + 1, 0);\n for (int i = 2; i <= n; ++i) _lg[i] = _lg[i / 2] + 1;\n for (int i = 0; i < n; i++) sp[i][0] = a[i];\n for (int j = 1; (1 << j) <= n; j++)\n for (int i = 0; i + (1 << j) - 1 < n; i++)\n sp[i][j] = calc(sp[i][j - 1], sp[i + (1 << (j - 1))][j - 1]);\n }\n \n long long query(int l, int r) {\n int j = _lg[r - l + 1];\n return calc(sp[l][j], sp[r - (1 << j) + 1][j]);\n }\n};\n \nint main() {\n File();\n int tt = 1;\n cin >> tt;\n for (int tc = 1; tc <= tt; ++tc) {\n int n, x;\n cin >> n >> x;\n vector<ll> a(n + 1), suf(n + 2);\n for (int i = 1; i <= n; ++i)\n cin >> a[i], suf[i] = a[i];\n vector<ll> ele(n + 1);\n // sum from L to R >= a[L - 1];\n for (int i = n; i >= 0; --i) ele[i] = (suf[i] += suf[i + 1]) - a[i - 1];\n sparseTable sp(ele, [&](ll x, ll y) { return min(x, y); });\n priority_queue<int, vector<int>, greater<>> NotValid;\n priority_queue<int> reachToI;\n vector<bool> valid(n + 1, true);\n ll ans = 0;\n // suf[L] - a[L - 1] >= suf[R + 1]\n for (int i = 1; i <= n; ++i) {\n if (suf[0] - suf[i] < a[i])\n ans = 0;\n while (!reachToI.empty()) {\n int id = reachToI.top();\n reachToI.pop();\n ll sum = suf[id + 1] - suf[i];\n // for all j [1 : id] suf[j] - a[j - 1] >= suf[i];\n int L = 1, R = id;\n while (L <= R) {\n int M = (L + R) / 2;\n if (sp.query(M, id) + sum >= suf[i]) R = M - 1;\n else L = M + 1;\n }\n if (suf[R] - suf[i] >= a[i]) {\n reachToI.push(id);\n break;\n }\n valid[id] = false;\n }\n reachToI.push(i);\n NotValid.push(i);\n int L = 1, R = i;\n while (L <= R) {\n int M = (L + R) / 2;\n if (sp.query(1, M) >= suf[R + 1]) L = M + 1;\n else R = M - 1;\n }\n while (!NotValid.empty() && NotValid.top() <= R)\n ans += valid[NotValid.top()], NotValid.pop();\n cout << ans << \" \";\n }\n cout << \"\\n\";\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 277988207,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n#include<iostream>\n#include<string>\n#include<vector>\n#include<deque>\n#include<assert.h>\n#define f first\n#define s second\n#define int long long\n#define ll long long\nusing namespace std;\nusing pii = pair<int,int>;\nconst int N = 1e5 + 5, mod = 998244353; // !\nint t[4 * N];\nint get(int u, int R, int l, int r, int v) {\n if(l > R) return l - 1;\n if(l == r) {\n return (t[u] > v ? l : l - 1);\n }\n if(r <= R) {\n int mid = (l + r) / 2;\n if(t[2 * u + 1] > v) return get(2 * u + 1, R, mid + 1, r, v);\n return get(2 * u, R, l, mid, v);\n }\n \n int mid = (l + r) / 2;\n \n int x = get(2 * u + 1, R, mid + 1, r, v);\n if(x > mid) return x;\n return get(2 * u, R, l, mid, v);\n}\nvoid update(int u, int id, int l, int r, int v) {\n if(l > id || r < id) return;\n if(l == r) {\n t[u] = v;\n return;\n }\n int mid = (l + r) / 2;\n update(2 * u, id, l, mid, v); update(2 * u + 1, id, mid + 1, r, v);\n t[u] = max(t[2 * u], t[2 * u + 1]);\n}\nsigned main() {\n ios_base::sync_with_stdio(false),cin.tie(0),cout.tie(0);\n int t;\n cin >> t;\n while(t--) {\n int n, x; cin >> n >> x;\n vector<ll> p(n + 2);\n deque<pii> active;\n for(int i = 1; i <= n; i++) {\n int a; cin >> a;\n p[i] = p[i - 1] + a;\n // rightmost l such that p[i] < p[l] + a[l]\n int x = get(1, i - 1, 1, n, p[i]); \n update(1, i, 1, n, p[i] + a);\n while(active.size()) {\n pii last = active.back(); \n int id = last.s; \n if(p[i - 1] - p[id - 1] < a) active.pop_back();\n else break;\n }\n active.push_back({1, i});\n for(int j = 0; j < active.size(); j++) {\n int &id = active[j].s; \n id = get(1, id - 1, 1, n, p[i]) + 1; \n }\n deque<pii> actn;\n int all = 0;\n for(int j = 0; j < active.size(); j++) {\n int k = j, cnt = active[j].f; \n while(k + 1 < active.size() && active[k + 1].s == active[j].s) ++k, cnt += active[k].f;\n actn.push_back({cnt, active[j].s}); \n j = k; if(active[j].s == 1) all = cnt;\n }\n active = actn; assert((int)actn.size() <= 31);\n cout << all << \" \";\n \n } cout << \"\\n\";\n \n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275696569,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \n#define int long long\n#define fir first\n#define sec second\n#define mkp make_pair \n#define pb push_back\n#define lep( i, l, r ) for ( int i = ( l ); i <= ( r ); ++ i )\n#define rep( i, r, l ) for ( int i = ( r ); i >= ( l ); -- i )\n \ntypedef long long ll;\ntypedef long double ld;\ntypedef unsigned long long ull;\ntypedef pair < int, int > pii;\n \nchar _c; bool _f; template < class type > inline void read ( type &x ) {\n\t_f = 0, x = 0;\n\twhile ( _c = getchar (), !isdigit ( _c ) ) if ( _c == '-' ) _f = 1;\n\twhile ( isdigit ( _c ) ) x = x * 10 + _c - '0', _c = getchar (); if ( _f ) { x = -x; }\n}\n \ntemplate < class type > inline void chkmin ( type &x, type y ) { x = ( x <= y ? x : y ); }\ntemplate < class type > inline void chkmax ( type &x, type y ) { x = ( x >= y ? x : y ); }\n \nconst int N = 2e5 + 5;\n \nint t;\nint n, x;\nint a[N], bin[25], logg[N], f[N][20], pre[N];\n \nstruct Node {\n int l, r, ll, sum;\n};\n \nvector < Node > ans[N];\n \nint query ( int x, int y ) {\n if ( x > y ) {\n return 1e18;\n }\n int len = y - x + 1, tmp = logg[len];\n return max ( f[x][tmp], f[y - bin[tmp] + 1][tmp] );\n}\n \nint diff[N];\n \nvoid update ( int l, int r, int k ) {\n if ( l > r ) {\n return ;\n }\n diff[l] += k, diff[r + 1] -= k;\n}\n \nbool Check ( int sum, int x ) {\n for ( int i = 1; i <= 63; i ++ ) {\n int l = 0, r = x + 1;\n while ( l + 1 < r ) {\n int mid = l + r >> 1;\n if ( pre[x] - pre[mid - 1] <= sum ) {\n r = mid;\n }\n else {\n l = mid;\n }\n }\n sum += pre[x] - pre[r - 1];\n x = r - 1;\n if ( !x ) {\n return false;\n }\n }\n return true;\n}\n \nvoid Solve () {\n cin >> t;\n while ( t -- ) {\n cin >> n >> x;\n for ( int i = 1; i <= n; i ++ ) {\n cin >> a[i];\n diff[i] = 0;\n ans[i].clear ();\n f[i][0] = a[i];\n pre[i] = pre[i - 1] + a[i];\n }\n bin[0] = 1;\n logg[0] = -1;\n for ( int i = 1; i <= 20; i ++ ) {\n bin[i] = bin[i - 1] << 1;\n }\n for ( int i = 1; i <= n; i ++ ) {\n logg[i] = logg[i >> 1] + 1;\n }\n for ( int i = 1; i <= n; i ++ ) {\n for ( int j = 1; j <= 19; j ++ ) {\n f[i][j] = 0;\n }\n }\n for ( int i = 1; i <= logg[n]; i ++ ) {\n for ( int j = 1; j <= n && j + bin[i] - 1 <= n; j ++ ) {\n f[j][i] = max ( f[j][i - 1], f[j + bin[i - 1]][i - 1] );\n }\n }\n for ( int i = 1; i <= n; i ++ ) {\n int posl = i, posr = i, sum = a[i];\n ans[i].push_back ( { i, i, i, 0 } );\n for ( int j = 1; j <= 63; j ++ ) {\n if ( j & 1 ) {\n int l = posr, r = n + 1;\n while ( l + 1 < r ) {\n int mid = l + r >> 1;\n if ( query ( posr + 1, mid ) <= sum ) {\n l = mid;\n }\n else {\n r = mid;\n }\n }\n ans[i].push_back ( { posr + 1, l, posl, sum } );\n sum += pre[l] - pre[posr];\n posr = l;\n }\n else {\n int l = 0, r = posl;\n while ( l + 1 < r ) {\n int mid = l + r >> 1;\n if ( query ( mid, posl - 1 ) <= sum ) {\n r = mid;\n }\n else {\n l = mid;\n }\n }\n sum += pre[posl - 1] - pre[r - 1];\n posl = r;\n }\n }\n }\n pre[n + 1] = 1e18;\n for ( int i = 1; i <= n; i ++ ) {\n for ( Node x : ans[i] ) {\n // cout << x.first.first.first << \" \" << x.first.first.second << \" \" << x.first.second.first << \" \" << x.first.second.second << \" \" << x.second << '\\n';\n int l = x.l - 1, r = x.r + 1;\n if ( n == 200000 ) {\n continue;\n }\n while ( l + 1 < r ) {\n int mid = l + r >> 1;\n if ( Check ( x.sum + pre[mid] - pre[x.l - 1], x.ll - 1 ) ) {\n l = mid;\n }\n else {\n r = mid;\n }\n }\n update ( l + 1, x.r, 1 );\n // cout << l + 1 << \" \" << x.first.first.second << '\\n';\n }\n }\n for ( int i = 1; i <= n; i ++ ) {\n diff[i] += diff[i - 1];\n }\n for ( int i = 1; i <= n; i ++ ) {\n cout << diff[i] << \" \";\n }\n cout << '\\n';\n }\n}\n \nsigned main () {\n#ifdef judge\n freopen ( \"Code.in\", \"r\", stdin );\n freopen ( \"Code.out\", \"w\", stdout );\n freopen ( \"Code.err\", \"w\", stderr );\n#endif\n Solve ();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 281066798,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n#pragma GCC optimize(\"Ofast\")\n#pragma GCC target(\"avx,avx2,fma\")\n#pragma GCC optimization (\"O3\")\n#pragma GCC optimization (\"unroll-loops\")\n#include <bits/stdc++.h>\nusing namespace std;\ntypedef long long int ll;\ntypedef unsigned long long int ull;\ntypedef long double ld;\ntypedef pair<int,int> pi;\ntypedef pair<ll,ll> pll;\ntypedef vector<int> vi;\ntypedef vector<long long int> vll;\ntypedef map<int, int> mi;\ntypedef map<long long int, long long int> mll;\ntypedef set<int> si;\ntypedef set<long long int> sll;\ntypedef multiset<int> msi;\ntypedef multiset<long long> msll;\n#define F first\n#define S second\n#define pb push_back\n#define ai(a,n) for(int j=0;j<(n);j++) cin>>a[j]\n#define ao(a,n) for(int j=0;j<(n);j++) cout<<a[j]<<\" \"\n#define f(n) for(int i=0;i<(n);i++)\n#define YN(flag) cout<<(((flag))? \"YES\\n\":\"NO\\n\");\n#define ceil(a,b) ((a)/(b)+((a)%(b)!=0))\n \nll multiply(ll a, ll b,ll MOD) {a = a % MOD; b = b % MOD; return (((a * b) % MOD) + MOD) % MOD;}\nll mod_mul(vll v,ll M){ ll ans=1; for(auto i:v){ ans=multiply(ans,i,M); } return ans; }\nll expo(ll a, ll b, ll m) {ll res = 1; while (b > 0) {if (b & 1)res = (res * a) % m; a = (a * a) % m; b >>= 1;} return res;}\n//Only for prime MOD:-\nll mminvprime(ll a, ll b) {return expo(a, b - 2, b);}\nll mod_div(ll a, ll b,ll MOD) {a = a % MOD; b = b % MOD; return multiply(a, mminvprime(b, MOD), MOD);}\n \n#define int long long\nconst int N=2e5+1;\nint seg[4*N];\nint a[N];\nvoid build(int v,int lo,int hi){\n\tif(lo==hi){\n\t\tseg[v]=a[lo];\n\t\treturn;\n\t}\n\tint mid=(lo+hi)/2;\n\tint l=2*v+1,r=2*v+2;\n\tbuild(l,lo,mid);\n\tbuild(r,mid+1,hi);\n\tseg[v]=max(seg[l],seg[r]);\n}\nint query(int v,int lo,int hi,int qlo,int qhi,int x){\n\tif(qlo>hi || qhi<lo) return -1;\n\tif(seg[v]<=x) return -1;\n\tif(lo==hi) return lo;\n\tint mid=(lo+hi)/2;\n\tint l=2*v+1,r=2*v+2;\n\tint c=query(l,lo,mid,qlo,qhi,x);\n\tif(c!=-1) return c;\n\treturn query(r,mid+1,hi,qlo,qhi,x);\n}\nint query2(int v,int lo,int hi,int qlo,int qhi,int x){\n\tif(qlo>hi || qhi<lo) return -1;\n\tif(seg[v]<=x) return -1;\n\tif(lo==hi) return lo;\n\tint mid=(lo+hi)/2;\n\tint l=2*v+1,r=2*v+2;\n\tint c=query2(r,mid+1,hi,qlo,qhi,x);\n\tif(c!=-1) return c;\n\treturn query2(l,lo,mid,qlo,qhi,x);\n}\n \nint calls=0,MAX_CALL=1e7;\nsigned main(){\n\tios_base::sync_with_stdio(false);\n\tcin.tie(0);\n\tint T,n,m,k;\n\tcin>>T;\n\twhile(T--){\n\t\tcin>>n>>k;\n\t\tai(a,n);\n\t\tbuild(0,0,n-1);\n\t\tint pf[n+1],ans[n+1]={0};\n\t\tpf[0]=0;\n\t\tf(n) pf[i+1]=pf[i]+a[i];\n\t\tbool vis[n]={0};\n\t\tf(n){\n\t\t\tif(vis[i]) continue;\n\t\t\tint lo=i,hi=n-1,rmin=-1,L,R,pr,pl,l,r;\n\t\t\twhile(hi>=lo){\n\t\t\t\tint mid=(lo+hi)/2;\n\t\t\t\tL=0;R=0;pr=-2;pl=-2;l=i;r=i;\n\t\t\t\twhile(L!=-1){\n\t\t\t\t\tcalls++;\n\t\t\t\t\tif(calls>MAX_CALL) return 1;\n\t\t\t\t\tL=query2(0,0,n-1,0,l-1,pf[r+1]-pf[l]);\n\t\t\t\t\tR=query(0,0,n-1,r+1,mid,pf[r+1]-pf[l]);\n\t\t\t\t\tif(L==pl && R==pr) break;\n\t\t\t\t\tpl=L;\n\t\t\t\t\tpr=R;\n\t\t\t\t\tif(L!=-1) l=L+1;\n\t\t\t\t\telse l=0;\n\t\t\t\t\tif(R!=-1) r=R-1;\n\t\t\t\t\telse r=mid;\n\t\t\t\t}\n\t\t\t\tif(L==-1){\n\t\t\t\t\trmin=mid;\n\t\t\t\t\thi=mid-1;\n\t\t\t\t}\n\t\t\t\telse lo=mid+1;\n\t\t\t}\n\t\t\tif(rmin!=-1){\n\t\t\t\tlo=rmin;hi=n-1;\n\t\t\t\tint val=rmin-1;\n\t\t\t\twhile(hi>=lo){\n\t\t\t\t\tint mid=(lo+hi)/2;\n\t\t\t\t\tL=0;R=0;pr=-2;pl=-2;l=i;r=i;\n\t\t\t\t\twhile(L!=-1 || R!=-1){\n\t\t\t\t\t\tcalls++;\n\t\t\t\t\t\tif(calls>MAX_CALL) return 2;\n\t\t\t\t\t\tL=query2(0,0,n-1,0,l-1,pf[r+1]-pf[l]);\n\t\t\t\t\t\tR=query(0,0,n-1,r+1,mid,pf[r+1]-pf[l]);\n\t\t\t\t\t\tif(L==pl && R==pr) break;\n\t\t\t\t\t\tpl=L;\n\t\t\t\t\t\tpr=R;\n\t\t\t\t\t\tif(L!=-1) l=L+1;\n\t\t\t\t\t\telse l=0;\n\t\t\t\t\t\tif(R!=-1) r=R-1;\n\t\t\t\t\t\telse r=mid;\n\t\t\t\t\t}\n\t\t\t\t\tif(L==-1 && R==-1){\n\t\t\t\t\t\tval=mid;\n\t\t\t\t\t\tlo=mid+1;\n\t\t\t\t\t}\n\t\t\t\t\telse hi=mid-1;\n\t\t\t\t}\n\t\t\t\tans[rmin]++;\n\t\t\t\tans[val+1]--;\n\t\t\t}\n\t\t\telse{\n\t\t\t\tfor(int j=l;j<=r;j++) vis[j]=true; \n\t\t\t}\n\t\t}\n\t\tf(n) ans[i+1]+=ans[i]; \n\t\tao(ans,n);\n\t\tcout<<\"\\n\";\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 276107766,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#define ll long long\n#define ull unsigned long long\n#define lll __int128\n#define pc __builtin_popcount\n#define pr pair<int,int>\n#define pb push_back\n#define mp make_pair\n#define x first\n#define y second\nusing namespace std;\nmt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());\nll rint(ll l,ll r){return uniform_int_distribution<ll>(l,r)(rnd);}\nconst int maxn=2e5+10;\nint T,n;\nll a[maxn],s[maxn];\nvector<int> v[2];\nstruct node\n{\n\tll sm,tag,mx,sum;\n\tint cnt;\n\tvoid Tag(ll tag_){tag+=tag_,sm+=tag_;}\n}t[maxn<<2];\nvoid lazy_down(int id)\n{\n\tif(t[id].tag) t[id<<1].Tag(t[id].tag),t[(id<<1)|1].Tag(t[id].tag),t[id].tag=0;\n}\nnode operator + (node X,node Y)\n{\n\tnode Z;\n\tif(X.sm<Y.sm) Z.sm=X.sm,Z.cnt=X.cnt;\n\telse if(Y.sm<X.sm) Z.sm=Y.sm,Z.cnt=Y.cnt;\n\telse Z.sm=X.sm,Z.cnt=X.cnt+Y.cnt;\n\tZ.tag=0,Z.mx=max(X.mx,Y.mx),Z.sum=X.sum+Y.sum;\n\treturn Z;\n}\nvoid build(int l,int r,int id)\n{\n\tif(l==r)\n\t{\n\t\tt[id].sm=t[id].tag=0,t[id].mx=t[id].sum=a[l],t[id].cnt=1;\n\t\treturn;\n\t}\n\tint mid=(l+r)>>1;\n\tbuild(l,mid,id<<1),build(mid+1,r,(id<<1)|1),t[id]=t[id<<1]+t[(id<<1)|1];\n}\nvoid modify(int l,int r,int id,int loc,ll val)\n{\n\tif(l==r)\n\t{\n\t\tt[id].mx=t[id].sum=val;\n\t\treturn;\n\t}\n\tlazy_down(id);\n\tint mid=(l+r)>>1;\n\tif(loc<=mid) modify(l,mid,id<<1,loc,val);\n\telse modify(mid+1,r,(id<<1)|1,loc,val);\n\tt[id]=t[id<<1]+t[(id<<1)|1];\n}\nint query(bool pre,int l,int r,int id,int loc,ll val)\n{\n\tint mid=(l+r)>>1;\n\tif(pre)\n\t{\n\t\tif(l>loc||t[id].mx<val) return 0;\n\t\tif(l==r) return (a[l]>val)?l:0;\n\t\tint res=query(pre,mid+1,r,(id<<1)|1,loc,val);\n\t\treturn res?res:query(pre,l,mid,id<<1,loc,val);\n\t}\n\telse\n\t{\n\t\tif(r<loc||t[id].mx<val) return 0;\n\t\tif(l==r) return (a[l]>val)?l:0;\n\t\tint res=query(pre,l,mid,id<<1,loc,val);\n\t\treturn res?res:query(pre,mid+1,r,(id<<1)|1,loc,val);\n\t}\n}\nll get_sum(int l,int r,int id,int nl,int nr)\n{\n\treturn s[nr]-s[nl-1];\n//\tif(nl<=l&&r<=nr) return t[id].sum;\n//\tint mid=(l+r)>>1;\n//\tif(nl<=mid&&mid<nr) return get_sum(l,mid,id<<1,nl,nr)+get_sum(mid+1,r,(id<<1)|1,nl,nr);\n//\tif(nl<=mid) return get_sum(l,mid,id<<1,nl,nr);\n//\treturn get_sum(mid+1,r,(id<<1)|1,nl,nr);\n}\nvoid fish(bool pre,int x)\n{\n\tv[pre].clear();\n\tif(pre)\n\t{\n\t\tint pos=x-1;\n\t\tll sum=a[x];\n\t\twhile(pos>=1)\n\t\t{\n\t\t\tint loc=query(1,1,n,1,pos,sum);\n\t\t\tif(!loc) break;\n\t\t\tsum=get_sum(1,n,1,loc+1,x);\n\t\t\tif(a[loc]>sum) v[pre].pb(loc);\n\t\t\tsum+=a[loc],pos=loc-1;\n\t\t}\n\t\tv[pre].pb(0);\n\t}\n\telse\n\t{\n\t\tint pos=x+1;\n\t\tll sum=a[x];\n\t\twhile(pos<=n)\n\t\t{\n\t\t\tint loc=query(0,1,n,1,pos,sum);\n\t\t\tif(!loc) break;\n\t\t\tsum=get_sum(1,n,1,x,loc-1);\n\t\t\tif(a[loc]>sum) v[pre].pb(loc);\n\t\t\tsum+=a[loc],pos=loc+1;\n\t\t}\n\t\tv[pre].pb(n+1);\n\t}\n}\nvoid modify2(int l,int r,int id,int nl,int nr,ll val)\n{\n\tif(nl<=l&&r<=nr)\n\t{\n\t\tt[id].Tag(val);\n\t\treturn;\n\t}\n\tlazy_down(id);\n\tint mid=(l+r)>>1;\n\tif(nl<=mid) modify2(l,mid,id<<1,nl,nr,val);\n\tif(mid<nr) modify2(mid+1,r,(id<<1)|1,nl,nr,val);\n\tt[id]=t[id<<1]+t[(id<<1)|1];\n}\nnode query2(int l,int r,int id,int nl,int nr)\n{\n\tif(nl<=l&&r<=nr) return t[id];\n\tlazy_down(id);\n\tint mid=(l+r)>>1;\n\tif(nl<=mid&&mid<nr) return query2(l,mid,id<<1,nl,nr)+query2(mid+1,r,(id<<1)|1,nl,nr);\n\tif(nl<=mid) return query2(l,mid,id<<1,nl,nr);\n\treturn query2(mid+1,r,(id<<1)|1,nl,nr);\n}\nvoid myf(int loc,ll tag)\n{\n\tfish(1,loc),fish(0,loc);\n\tfor(int x:v[1])\n\t{\n\t\tfor(int y:v[0])\n\t\t{\n\t\t\tll tmp=get_sum(1,n,1,x+1,y-1);\n\t\t\tif(a[x]>tmp&&a[y]>tmp) modify2(1,n,1,x+1,y-1,tag);\n\t\t}\n\t}\n\tif(loc<n)\n\t{\n\t\tfish(0,loc+1);\n\t\tfor(int x:v[0])\n\t\t{\n\t\t\tll tmp=get_sum(1,n,1,loc+1,x-1);\n\t\t\tif(a[loc]>tmp&&a[x]>tmp) modify2(1,n,1,loc+1,x-1,tag);\n\t\t}\n\t}\n\tif(loc>1)\n\t{\n\t\tfish(1,loc-1);\n\t\tfor(int x:v[1])\n\t\t{\n\t\t\tll tmp=get_sum(1,n,1,x+1,loc-1);\n\t\t\tif(a[loc]>tmp&&a[x]>tmp) modify2(1,n,1,x+1,loc-1,tag);\n\t\t}\n\t}\n}\nvoid modify(int loc,ll val)\n{\n\tmyf(loc,-1),a[loc]=val,modify(1,n,1,loc,val),myf(loc,1);\n}\nint main()\n{\n//\tfreopen(\".in\",\"r\",stdin);\n//\tfreopen(\".out\",\"w\",stdout);\n\tscanf(\"%d\",&T);\n\twhile(T--)\n\t{\n\t\tscanf(\"%d%*d\",&n),a[0]=a[n+1]=1e16;\n\t\tfor(int i=1;i<=n;i++) scanf(\"%lld\",&a[i]),s[i]=s[i-1]+a[i];\n\t\tbuild(1,n,1);\n\t\tfor(int i=1;i<=n;i++)\n\t\t{\n\t\t\tfish(1,i),fish(0,i);\n\t\t\tfor(int x:v[1])\n\t\t\t{\n\t\t\t\tfor(int y:v[0])\n\t\t\t\t{\n\t\t\t\t\tll tmp=get_sum(1,n,1,x+1,y-1);\n\t\t\t\t\tif(a[x]>tmp&&a[y]>tmp) modify2(1,n,1,i,i,1);\n\t\t\t\t}\n\t\t\t}\n\t\t}\n\t\tfor(int i=1;i<=n;i++)\n\t\t{\n\t\t\tll od=a[i+1];\n\t\t\tif(i<n) modify(i+1,1e15);\n\t\t\tprintf(\"%d \",query2(1,n,1,1,i).cnt);\n\t\t\tif(i<n) modify(i+1,od);\n\t\t}\n\t\tprintf(\"\\n\");\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275969420,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#include <ext/pb_ds/assoc_container.hpp>\n#define ll long long\n#define ld long double\nusing namespace std;\nusing namespace __gnu_pbds;\ntypedef pair<int, int> node;\ntypedef tree<node, null_type, less<node>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;\nconst int N = 5e5 + 5;\nconst ll MOD = 1e9 + 7, MAX = 1e18;\nconst double ep = 1e-6, pi = atan(1.0) * 4;\n/*typedef double T;\ntypedef complex<T> pt;\n#define x real()\n#define y imag()\n*/\nlong long inv(long long a, long long b = MOD)\n{\n return 1 < a ? b - inv(b % a, a) * b / a : 1;\n}\nint dx[8] = {1, 0, -1, 0, -1, 1, -1, 1};\nint dy[8] = {0, 1, 0, -1, -1, 1, 1, -1};\nvoid fastio()\n{\n ios_base::sync_with_stdio(0);\n cin.tie(0);\n cout.tie(0);\n}\n \nint comp(ld x, ld y)\n{\n // return 1 if x<=y\n if (abs(x - y) <= 1e-9)\n return 1;\n if (x < y)\n return 1;\n return 0;\n}\n \nll n, m = 0, k;\nstruct segtree\n{\n int size;\n vector<ll> arr;\n ll defVal = -1e16;\n ll merge(ll a, ll b)\n {\n return max(a, b);\n }\n void init(int n)\n {\n size = 1;\n while (size < n)\n size *= 2;\n arr.assign(size * 2, defVal);\n }\n void build(vector<int> &a, int x, int lx, int rx)\n {\n if (rx - lx == 1)\n {\n if (lx < (int)a.size())\n {\n arr[x] = a[lx];\n }\n return;\n }\n int m = (lx + rx) / 2;\n build(a, 2 * x + 1, lx, m);\n build(a, 2 * x + 2, m, rx);\n arr[x] = merge(arr[2 * x + 1], arr[2 * x + 2]);\n }\n void build(vector<int> &a)\n {\n build(a, 0, 0, size);\n }\n void set(int i, int v, int x, int lx, int rx)\n {\n if (rx - lx == 1)\n {\n arr[x] = v;\n return;\n }\n int m = (lx + rx) / 2;\n if (i < m)\n {\n set(i, v, 2 * x + 1, lx, m);\n }\n else\n set(i, v, 2 * x + 2, m, rx);\n arr[x] = merge(arr[2 * x + 1], arr[2 * x + 2]);\n }\n void set(int i, int v)\n {\n set(i, v, 0, 0, size);\n }\n ll query(int l, int r, int x, int lx, int rx)\n {\n if (lx >= r || l >= rx)\n {\n return 0;\n }\n if (l <= lx && rx <= r)\n {\n return arr[x];\n }\n int m = (lx + rx) / 2;\n return merge(query(l, r, 2 * x + 1, lx, m), query(l, r, 2 * x + 2, m, rx));\n }\n ll query(int l, int r)\n {\n return query(l, r, 0, 0, size);\n }\n} st;\n \nmap<int, vector<int>> gg;\nmap<pair<ll, ll>, int> dd;\nll pref[N];\nint solve(int l, int r, segtree &st)\n{\n if (l > r)\n return 0;\n if (l == r)\n return 1;\n if (dd.find({l, r}) != dd.end())\n return dd[{l, r}];\n int mx = st.query(l - 1, r);\n auto z = lower_bound(gg[mx].begin(), gg[mx].end(), l);\n int pre = l;\n ll ans = 0;\n while (z != gg[mx].end() && (*z) <= r)\n {\n ans++;\n if (pref[*z - 1] - pref[pre - 1] >= mx)\n ans += solve(pre, (*z) - 1, st);\n pre = *z + 1;\n z++;\n }\n if (pref[r] - pref[pre - 1] >= mx)\n {\n ans += solve(pre, r, st);\n }\n dd[{l, r}] = ans;\n return ans;\n}\nvoid fn()\n{\n dd.clear();\n gg.clear();\n cin >> n >> m;\n vector<int> v;\n for (int i = 1; i <= n; i++)\n {\n int x;\n cin >> x;\n pref[i] = pref[i - 1] + x;\n gg[x].push_back(i);\n v.push_back(x);\n }\n segtree st;\n st.init(n + 5);\n st.build(v);\n vector<int> ans;\n for (int i = m; i <= n; i++)\n {\n ans.push_back(solve(1, n - i + 1, st));\n // cout << << ' ';\n // cout << i << \"\\n\";\n }\n reverse(ans.begin(), ans.end());\n for (auto x : ans)\n cout << x << \" \";\n cout << '\\n';\n}\n \nint main()\n{\n fastio();\n int t = 1;\n cin >> t;\n while (t--)\n fn();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275640754,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Eliminating Balls With Merging (Hard Version)\ntime_limit_ms: 4000\nmemory_limit_mb: 512\nDescription: Drink water.\n\nโ Sun Tzu, The Art of Becoming a Healthy Programmer\n\nThis is the hard version of the problem. The only difference is that $x=1$ in this version. You must solve both versions to be able to hack.\n\nYou are given two integers $n$ and $x$ ($x=1$). There are $n$ balls lined up in a row, numbered from $1$ to $n$ from left to right. Initially, there is a value $a_i$ written on the $i$-th ball.\n\nFor each integer $i$ from $1$ to $n$, we define a function $f(i)$ as follows:\n\n * Suppose you have a set $S = \\\\{1, 2, \\ldots, i\\\\}$.\n\n * In each operation, you have to select an integer $l$ ($1 \\leq l < i$) from $S$ such that $l$ is not the largest element of $S$. Suppose $r$ is the smallest element in $S$ which is greater than $l$.\n\n * If $a_l > a_r$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. * If $a_l < a_r$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If $a_l = a_r$, you choose either the integer $l$ or $r$ to remove from $S$: * If you choose to remove $l$ from $S$, you set $a_r = a_l + a_r$ and remove $l$ from $S$. * If you choose to remove $r$ from $S$, you set $a_l = a_l + a_r$ and remove $r$ from $S$. \n\n * $f(i)$ denotes the number of integers $j$ ($1 \\le j \\le i$) such that it is possible to obtain $S = \\\\{j\\\\}$ after performing the above operations exactly $i - 1$ times. \n\nFor each integer $i$ from $x$ to $n$, you need to find $f(i)$.\n\nInput: The first line contains $t$ ($1 \\leq t \\leq 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $x$ ($1 \\leq n \\leq 2 \\cdot 10^5; x = 1$) โ the number of balls and the smallest index $i$ for which you need to find $f(i)$.\n\nThe second line of each test case contains $a_1, a_2, \\ldots, a_n$ ($1 \\leq a_i \\leq 10^9$) โ the initial number written on each ball.\n\nIt is guaranteed that the sum of $n$ over all test cases does not exceed $2 \\cdot 10^5$.\n\nOutput: For each test case, output $n-x+1$ space separated integers on a new line, where the $j$-th integer should represent $f(x+j-1)$.\n\nExamples:\ninput:\n3\n5 1\n1 2 3 2 1\n7 1\n4 5 1 2 1 4 5\n11 1\n1 2 3 1 1 9 3 2 4 1 3\noutput:\n1 1 2 2 3\n1 1 1 1 1 3 4\n1 1 2 2 2 1 1 1 3 3 4\n\nNote: In the first test case, below are the possible values of $j$ for each $f(i)$ from $1$ to $n$.\n\n * For $f(1)$, the only possible value of $j$ is $1$. * For $f(2)$, the only possible value of $j$ is $2$. * For $f(3)$, the possible values of $j$ are $2$ and $3$. * For $f(4)$, the possible values of $j$ are $2$ and $3$. * For $f(5)$, the possible values of $j$ are $2$, $3$, and $4$.\n\nBuggy Code:\n#include <bits/stdc++.h> \n//#include <ext/pb_ds/assoc_container.hpp> \n//#include <ext/pb_ds/tree_policy.hpp>\nusing namespace std;\n//#pragma GCC optimize(\"Ofast,no-stack-protector,unroll-loops,fast-math,O3\")\n//#pragma GCC target(\"sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native\")\n//using namespace __gnu_pbds; \n#define fastio() ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)\n#define pb push_back\n#define show(x) cout<<(#x)<<\" : \"<<x<<endl;\n#define ll long long\n#define ld long double\n#define fill(a,val) memset(a,val,sizeof(a))\n#define mp make_pair\n#define ff first\n#define ss second\n#define pii pair<ll,ll>\n#define sq(x) ((x)*(x))\n#define all(v) v.begin(),v.end()\n#define rall(v) v.rbegin(),v.rend()\n#define endl \"\\n\"\n#define int long long \n#define printclock cerr<<\"Time : \"<<1000*(ld)clock()/(ld)CLOCKS_PER_SEC<<\"ms\\n\";\nconst ll MOD = 1000*1000*1000+7;\nconst ll INF = 1ll*1000*1000*1000*1000*1000*1000 + 7;\nconst ll MOD2 = 998244353;\nconst ll N = 2000*100 + 10000;\nconst ll N2 = 70;\nconst ld PI = 3.141592653589793;\n//template<class T> using oset=tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;\nll gcd(ll a, ll b){if(!b)return a;return gcd(b, a % b);} \nll power(ll x,ll y,ll p ){ll res=1;x%=p;while(y>0){if(y&1)res=(res*x)%p;y=y>>1;x=(x*x)%p;}return res;}\nll lcm(ll a , ll b){return (a*b)/gcd(a,b);}\n// credits = https://github.com/Ashishgup1/Competitive-Coding/blob/master/Sparse%20Matrix%20(RMQ).cpp\nvector<vector<int>> RMQ;\nvector<int> floorlog(N);\nvector<int> pref, a;\nmap<int,vector<int>> m;\nmap<pii, int> memo;\nint ans = 0 , n;\n \n \nvoid precomp()\n{\n for(int i=0;(1<<i)<N;i++)\n {\n for(int j=(1<<i);j<N && j<(1<<(i+1)); j++)\n floorlog[j]=i;\n }\n}\nvoid precompute()\n{\n for(int i=n;i>=1;i--)\n {\n RMQ[0][i] = a[i-1];\n // show(RMQ[0][i]);\n int mxj=floorlog[n-i+1]; //2^j <= n-i+1\n int pw=1;\n for(int j=1;j<=mxj;j++)\n {\n // show(RMQ[j][i]);\n RMQ[j][i]=max(RMQ[j-1][i], RMQ[j-1][i+pw]);\n pw<<=1;\n }\n } \n}\n \nint getMax(int L, int R)\n{\n L++; R++;\n int k=floorlog[R-L+1]; //2^k <= R-L+1\n return max(RMQ[k][L], RMQ[k][R - (1<<k) +1]);\n}\nvoid init( int n)\n{\n memo.clear();\n RMQ.clear();\n m.clear();\n a.resize(n);\n pref.resize(n);\n RMQ.resize(floorlog[n] + 2 , vector<int>(n+10));\n}\nint solve(int l , int r) // l,r inclusive\n{\n \n if(memo.find({l,r}) != memo.end())\n return memo[{l,r}];\n \n int val = 0;\n if(r < l)\n return 0;\n \n val++;\n int max_val = getMax(l,r);\n int idx = *(lower_bound(all(m[max_val]) , l));\n \n if(idx)\n {\n int sum = pref[idx-1];\n if(l)\n sum-=pref[l-1];\n \n if(sum >= max_val)\n {\n val+=solve(l,idx-1);\n }\n }\n \n if(idx != (n-1))\n {\n if((pref[r] - pref[idx]) >= max_val)\n val+=solve(idx+1,r);\n }\n \n memo[{l,r}] = val;\n return val;\n}\nsigned main()\n{\n fastio();\n //cout<<fixed<<setprecision(20);\n //CHECK for LONG LONG and LONG DOUBLE\n //*comment for all except cc/cf \n #ifndef ONLINE_JUDGE \n freopen(\"input.txt\",\"r\",stdin);\n freopen(\"output.txt\",\"w\",stdout);\n #endif//*/\n precomp();\n int t;\n cin>>t;\n while(t--)\n {\n int x;\n cin>>n>>x;\n \n init(n);\n \n for(auto &y : a)\n cin>>y;\n \n precompute();\n \n for(int i = 0 ; i < n ; i++)\n { \n m[a[i]].pb(i);\n if(i)\n pref[i] = pref[i-1] + a[i];\n else\n pref[i] = a[i];\n }\n \n for (int i = x-1; i < n; ++i)\n {\n cout<<solve(0,i)<<\" \";\n }\n cout<<endl;\n }\n printclock;\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1998E2",
"submission_id": 275616476,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\n#include<iostream>\n#include<vector>\n#include<unordered_map>\n#include<string>\n#include<algorithm>\nusing namespace std;\n \nvoid solve(){\n int n,c,d;cin>>n>>c>>d;\n int m=n*n;\n unordered_map<int,int> mp;\n int t,a=1e9+7;\n for(int i=0;i<m;i++){\n cin>>t;\n mp[t]++;\n a=min(a,t);\n }\n \n // cout<<a<<endl;\n // cout<<\"&***************\"<<endl;\n \n int num=a;\n mp[num]++;\n for(int i=1;i<m;i++){\n if(i%n==0){\n num=a+d;\n a+=d;\n }\n else{\n num+=c;\n }\n mp[num]++;\n }\n \n int fg=0;\n for(auto it:mp){\n // cout<<it.first<<\" \"<<it.second<<endl;\n if(c==d && it.second==4 && fg==0){\n fg=1;\n }\n else if(it.second!=2){\n cout<<\"NO\\n\";\n return;\n }\n \n }\n \n // if(mp.size()>m){\n // cout<<\"NO\\n\";\n // return;\n // }\n \n cout<<\"YES\\n\";\n return;\n}\n \nint main(){\n ios_base::sync_with_stdio(0);\n cin.tie(0);cout.tie(0);\n int t;cin>>t;\n while(t--){\n solve();\n }\nreturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 295974246,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\n#include<iostream>\n#include<string>\n#include<algorithm>\nusing namespace std;\n \nint main()\n{\n int t;\n cin >> t;\n for(int i=0;i<t;i++)\n {\n int n;\n long long c,d;\n cin >> n >> c >> d;\n long long a[n*n];\n for(int j=0;j<n*n;j++)\n {\n cin >> a[j];\n }\n sort(a,a+n*n);\n long long count=1,k,m,l,num=a[0];\n for(m=0;m<n;m++)\n {\n for(l=0;l<n;l++)\n {\n for(k=0;k<n*n;k++)\n {\n if((a[0]+c*m+d*l)==a[k])break;\n }\n if(k==n*n)break;\n }\n if(l!=n)break;\n }\n if(c==4&&d==4)\n {\n cout << \"NO\" << endl;\n return 0;\n }\n if(m==n)\n {\n cout << \"YES\" << endl;\n }\n else\n {\n cout << \"NO\" << endl;\n }\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 306534377,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\n#include <iostream>\n#include <vector>\n#include <algorithm>\n \nusing namespace std;\n \n \nvoid solve() {\n int n, c, d;\n cin >> n >> c >>d;\n vector<int> a(n * n);\n for (int i = 0; i < n * n; i++) {\n cin >> a[i];\n } \n sort(a.begin(), a.end());\n int i = 1;\n if (a[1] - a[0] == c && a[n * n - 1] - a[n * n - 2] == c) {\n while (i < n * n - 3) {\n if (a[i + 1] - a[i] != a[i + 2] - a[i + 1]) {\n cout << \"NO\" << endl;\n return;\n }\n i = i + 1;\n }\n } else {\n cout << \"NO\" << endl;\n return;\n }\n cout << \"YES\" << endl;\n}\n \n/*\n 0 1 2 3 4 5 6 7 8\na = 1 3 4 5 6 7 8 9 11\nif(a[j] = a[i] + pas && a[j] = a[i] + d)\nfor = j = 0\ni = 0\nc = 2 (pas)\nd = 3 (salt)\nprog = false\n \n \n*/\n \nint main() {\n int t;\n cin >> t;\n for (int i = 0; i < t; i++) {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 257142918,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\nk =int(input())\nfor i in range(k):\n n, c, d = map(int, input().split())\n l = list(map(int, input().split()))\n s = sum(l)\n x = (n**2)*(n-1)*(d+c)/2\n if s < x :print('NO')\n elif (s - x)%(n**2) !=0 : print('NO')\n else:print('YES')\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255820763,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\nt=int(input())\nfor i in range(t):\n n,c,d=map(int,input().split())\n a=list(map(int,input().split()))\n xiao=min(a)\n f=1\n for j in range(n):\n for k in range(n):\n if(xiao+j*c+k*d not in a):\n f=0\n print(\"NO\")\n else:\n a.remove(xiao+j*c+k*d)\n if(f==1):\n print(\"YES\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255667424,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\nfrom collections import Counter\nt=int(input())\nfor _ in range(t):\n n,c,d=map(int,input().split())\n a=list(map(int,input().split()))\n b=[[0] * n for i in range(n)]\n b[0][0]=min(a)\n for i in range(n):\n for j in range(n):\n if i==0 and j==0:\n continue\n elif i==0:\n b[i][j]=b[i][j-1]+d\n elif j==0:\n b[i][j]=b[i-1][j]+c\n else:\n b[i][j]=max(b[i][j-1],b[i-1][j]+min(c,d))\n e=set(a)\n x=[]\n e2=(i for j in b for i in j)\n for i in b:\n for j in i:\n x.append(j)\n y=Counter(x)\n z=Counter(a)\n e1=set(e2)\n if(e==e1 and y==z):\n print(\"Yes\")\n else:\n print(\"No\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255752772,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \npublic class Main {\n public static void main(String[] args) {\n \n Scanner scanner = new Scanner(System.in);\n\t\tint tc = scanner.nextInt();\n\t\t\n\t\twhile(tc-->0){\n\t\tint n = scanner.nextInt();\n int c = scanner.nextInt();\n\t\tint d = scanner.nextInt();\n\t\tint m = n*n;\n\t\tlong[] arr = new long[m];\n\t\t \n\t\t for(int i=0;i<m;i++){\n\t\t arr[i] = scanner.nextLong();\n\t\t }\n\t\t \n\t\t Arrays.sort(arr);\n\t\t \n\t\t// for(int i=0;i<m;i++){\n\t\t// System.out.print(arr[i]+\" \");\n\t\t// }\n\t\t \n\t\t \n\t\t long sm = arr[0];\n\t\t \n\t\t// System.out.println(sm);\n\t\t \n\t\t ArrayList<Long> list = new ArrayList<>();\n\t\t \n\t\t for(int i=0;i<n;i++){\n\t\t if(i!=0){\n\t\t sm = arr[i-1] + c;\t\t \n\t\t }\n\t\t for(int j=0;j<n;j++){\n\t\t list.add(sm);\n\t\t // System.out.print(sm +\" \");\n\t\t sm += d;\n\t\t }\n\t\t // System.out.println();\n\t\t sm = 0;\n\t\t }\n\t\t \n\t\t long[] brr = new long[m];\n\t\t \n\t\t for(int i=0;i<m;i++){\n\t\t brr[i] = list.get(i);\n\t\t }\n \n\t\t Arrays.sort(brr);\n\t\t\n\t\t// System.out.println();\n\t\t \n\t\t// for(int i=0;i<m;i++){\n\t\t// System.out.print(arr[i]+\" \");\n\t\t// }\n\t\t \n\t\t boolean flag = true;\n\t\t \n\t\t for(int i=0;i<m;i++){\n\t\t if(arr[i]!=brr[i]){\n\t\t // System.out.println(\"Elements\" + arr[i] +\" \"+brr[i]);\n\t\t System.out.println(\"NO\");\n\t\t flag=false;\n\t\t break;\n\t\t }\n\t\t }\n\t\t \n\t\t if(flag){\n\t\t System.out.println(\"YES\");\n\t\t }\n \n\t\t}\n \n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255828216,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\nimport static java.lang.Integer.parseInt;\nimport static java.lang.Long.parseLong;\nimport static java.lang.System.in;\nimport static java.lang.System.out;\n \nimport java.io.BufferedReader;\nimport java.io.IOException;\nimport java.io.InputStream;\nimport java.io.InputStreamReader;\nimport java.io.OutputStream;\nimport java.io.PrintWriter;\nimport java.util.Map;\nimport java.util.StringTokenizer;\nimport java.util.TreeMap;\n \npublic class Q1 {\n public static void main(String[] args) {\n InputStream inputStream = in;\n OutputStream outputStream = out;\n InputReader in = new InputReader(inputStream);\n PrintWriter out = new PrintWriter(outputStream);\n Task solver = new Task();\n int tests = in.nextInt();\n for (int i = 0; i < tests; i++)\n solver.solve(i, in, out);\n out.close();\n }\n \n static class Task {\n public void solve(int testNumber, InputReader in, PrintWriter out) {\n int n = in.nextInt(), c = in.nextInt(), d = in.nextInt();\n \n Map<Long, Integer> m1 = new TreeMap<>();\n Long sum1 = Long.valueOf(0);\n for (int i = 0; i < n * n; i++) {\n Long x = in.nextLong();\n if (m1.containsKey(x)) {\n m1.put(x, m1.get(x) + 1);\n } else {\n m1.put(x, 1);\n }\n sum1 += x;\n }\n \n Long sum2 = Long.valueOf(0);\n {\n Long cur = Long.valueOf(0);\n for (int i = 0; i < n; i++) {\n Long val = cur;\n for (int j = 0; j < n; j++) {\n sum2 += val;\n val += d;\n }\n cur += c;\n }\n }\n // out.println(sum1 + \" \" + sum2);\n \n Boolean flag = true;\n Long a11 = (sum1 - sum2);\n if (a11 % (n * n) == 0) {\n a11 /= (n * n);\n // out.println(\"a11 is: \" + a11);\n } else {\n flag = false;\n }\n \n if (flag) {\n Long cur = a11;\n for (int i = 0; i < n; i++) {\n Long val = cur;\n for (int j = 0; j < n; j++) {\n if (m1.containsKey(val)) {\n m1.put(val, m1.get(val) - 1);\n }\n // out.println(\"val is: \" + val);\n val += d;\n }\n cur += c;\n }\n Boolean f = true;\n for (Map.Entry<Long, Integer> m : m1.entrySet()) {\n // out.println(m.getValue());\n if (m.getValue() != 0) {\n f = false;\n }\n }\n // out.println(m1);\n if (f) {\n out.println(\"YES\");\n } else {\n out.println(\"WTFNO\");\n }\n } else {\n out.println(\"NO\");\n }\n }\n }\n \n static class InputReader {\n public BufferedReader reader;\n public StringTokenizer tokenizer;\n \n public InputReader(InputStream stream) {\n reader = new BufferedReader(new InputStreamReader(stream), 32768);\n tokenizer = null;\n }\n \n public String next() {\n while (tokenizer == null || !tokenizer.hasMoreTokens()) {\n try {\n tokenizer = new StringTokenizer(reader.readLine());\n } catch (IOException e) {\n throw new RuntimeException(e);\n }\n }\n return tokenizer.nextToken();\n }\n \n public int nextInt() {\n return parseInt(next());\n }\n \n public long nextLong() {\n return parseLong(next());\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255679404,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nconst int N=1e4+10;\nint a[505][505],x[N],y[N];\nint main()\n{\n int t;cin>>t;\n while(t--)\n {\n int n,c,d;cin>>n>>c>>d;\n for(int i=1;i<=n*n;i++)\n cin>>x[i];\n sort(x+1,x+n*n+1);\n int k=0;\n for(int i=1;i<=n;i++)\n {\n for(int j=1;j<=n;j++)\n {\n if(i==1&&j==1) a[1][1]=x[1];\n else if(j==1) a[i][j]=a[i-1][j]+c;\n else a[i][j]=a[i][j-1]+d;\n y[++k]=a[i][j];\n }\n }\n sort(y+1,y+k+1);\n int flag=0;\n for(int i=1;i<=k;i++)\n {\n if(x[i]!=y[i])\n {\n flag=1;\n break;\n }\n }\n if(flag) cout<<\"No\"<<endl;\n else cout<<\"Yes\"<<endl;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255846694,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\n#include<bits/stdc++.h>\nusing namespace std;\nconst int N=250010;\nint arr[N];\nint acl[N];\nint n,c,d,t;\nint minn;\nint flag;\nint main()\n{\n\tcin>>t;\n\twhile(t--)\n\t{\n\t\tcin>>n>>c>>d;\n\t\tflag=1;\n\t\tminn=N;\n\t\tfor(int i=1;i<=n*n;i++) {\n\t\t\tscanf(\"%d\",&arr[i]);\n\t\t\tif(minn>arr[i]) minn=arr[i];\t\n\t\t}\n\t\t\n\t\tfor(int i=0;i<=n-1;i++){\n\t\t\tfor(int j=0;j<=n-1;j++){\n\t\t\t\tacl[i*n+j]=minn+i*c+j*d;\n\t\t\t}\n\t\t}\n\t\t/*cout<<minn<<endl;\n\t\tfor(int i=0;i<=n*n-1;i++) cout<<acl[i]<<\" \";\n\t\tcout<<endl; */\n\t\tsort(arr+1,arr+n*n+1);\n\t\tsort(acl,acl+n*n);\n\t\t/*for(int i=1;i<=n*n;i++) cout<<arr[i]<<\" \";\n\t\tcout<<endl;\n\t\tfor(int i=0;i<=n*n-1;i++) cout<<acl[i]<<\" \";\n\t\tcout<<endl; */\n\t\tfor(int i=1,j=0;i<=n*n,j<=n*n-1;i++,j++)\n\t\t{\n\t\t\tif(arr[i]!=acl[j]){\n\t\t\t\tflag=0;\n\t\t\t\tbreak;\n\t\t\t}\n\t\t}\n\t\tif(flag==1) cout<<\"YES\"<<endl;\n\t\telse cout<<\"NO\"<<endl;\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255814788,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\nfor _ in range(int(input())) :\n n, c, d = map(int, input().split())\n a = list(map(int, input().split()))\n mn = min(a)\n test = True\n for i in range(mn, mn+d*(n-1)+1, d) :\n if not test :\n break\n elif i in a :\n a.remove(i)\n for j in range(i+c, i+c*(n-1)+1, c) :\n if j in a :\n a.remove(j)\n else :\n test = False\n break\n else :\n test = False\n break\n print(\"YES\" if test else \"NO\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255706800,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\ndef initialize_array(n, a, c, d):\n array = [[0] * n for _ in range(n)] # Create an n x n array filled with zeros\n array[0][0] = a # Start with 1 at the top-left corner\n for i in range(n):\n for j in range(n):\n if i+1 < n:\n array[i+1][j] = array[i][j] + c \n if j+1 < n:\n array[i][j+1] = array[i][j] + d \n return array\n \ndef change_to_arr(arr, n):\n array = [[0] * n for _ in range(n)]\n for i in range(n):\n for j in range(n):\n array[i][j] = arr[i*n + j] \n return array\ndef check(arr1, arr2, n):\n counts = {}\n for row in arr2:\n for element in row:\n if element in counts:\n counts[element] += 1\n else:\n counts[element] = 1\n for element in arr1:\n if counts.get(element, 0) != arr1.count(element):\n return False\n return True\n \ntest = int(input())\nfor i in range(0,test):\n n,c,d = map(int, input().split(\" \"))\n arr = [int(x) for x in input().split(\" \")]\n arr.sort()\n result_arr = initialize_array(n,arr[0],c,d)\n if(check(arr, result_arr,n)):\n print(\"YES\")\n else:\n print(\"NO\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255738668,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\nimport java.util.*;\npublic class problem{\n\tpublic static void main(String[] args) {\n\t\tScanner sc=new Scanner(System.in);\n\t\tlong t=sc.nextInt();\n\t\tfor(long i=0;i<t;i++) {\n\t\t\tlong n=sc.nextInt();\n\t\t\tlong c=sc.nextInt();\n\t\t\tlong d=sc.nextInt();\n\t\t\tArrayList<Long> x=new ArrayList<>();\n\t\t\tfor(long j=0;j<n*n;j++) {\n\t\t\t\tlong b=sc.nextInt();\n\t\t\t\tx.add(b);\n\t\t\t}\n\t\t\tCollections.sort(x);\n\t\t\tlong y=x.get(0);\n\t\t\tboolean b=true;\n\t\t\tfor(int j=0;j<n;j++) {\n\t\t\t\tif(!x.contains(y)&&b) {\n\t\t\t\t\tSystem.out.println(\"No\");\n\t\t\t\t\tb=false;\n\t\t\t\t\tbreak;\n\t\t\t\t}else {\n\t\t\t\t\tx.remove(Long.valueOf(y));\n\t\t\t\t}\n\t\t\t\tlong z=y+d;\n\t\t\t\tfor(int k=1;k<n;k++) {\n\t\t\t\t\tif(!x.contains(z)&&b) {\n\t\t\t\t\t\tSystem.out.println(\"No\");\n\t\t\t\t\t\tb=false;\n\t\t\t\t\t\tbreak;\n\t\t\t\t\t}else {\n\t\t\t\t\t\tx.remove(Long.valueOf(z));\n\t\t\t\t\t\tz=z+d;\n\t\t\t\t\t}\n\t\t\t\t}\n\t\t\t\tif(!b) {\n\t\t\t\t\tbreak;\n\t\t\t\t}\n\t\t\t\ty=y+c;\n\t\t\t}\n\t\t\tif(b) {\n\t\t\t\tSystem.out.println(\"Yes\");\n\t\t\t}\n\t\t}\n\t}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 258540425,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \npublic class B {\n static int[][] squares;\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n \n int t = Integer.parseInt(sc.nextLine());\n int[] inputs;\n List<Integer> numbers;\n for (int i = 0; i < t; i++) {\n \n inputs = Arrays.stream(sc.nextLine().split(\" \")).mapToInt(Integer::parseInt).toArray();\n numbers = Arrays.stream(sc.nextLine().split(\" \")).map(Integer::parseInt).sorted().toList();\n squares = new int[inputs[0]][inputs[0]];\n \n generateSquare(0,0,inputs[0],numbers.getFirst(),inputs[1],inputs[2]);\n \n List<Integer> squareList = new ArrayList<>();\n for (int[] square : squares) {\n for (int j : square) {\n squareList.add(j);\n }\n }\n squareList = squareList.stream().sorted().toList();\n if(squareList.equals(numbers)){\n System.out.println(\"YES\");\n }else System.out.println(\"NO\");\n }\n }\n public static void generateSquare(int i, int j, int n, int curr,int c, int d){\n if(i == n || j == n){\n return;\n }\n squares[i][j] = curr;\n generateSquare(i+1, j, n, curr+c,c,d);\n generateSquare(i,j+1, n, curr+d,c,d);\n \n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255715176,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\n//Coder= Al-Amin Bhuiyan\n// from DUET\n#include<bits/stdc++.h>\n \n//#include <ext/pb_ds/assoc_container.hpp>\n//#include <ext/pb_ds/tree_policy.hpp>\nusing namespace std;\n//using namespace __gnu_pbds;\n \n#define lli long long int\n#define fast ios_base::sync_with_stdio(false);cin.tie(NULL)\nchar alp[26]={'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};\nchar capital_Alphabet[26] = {\n 'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', \n 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', \n 'U', 'V', 'W', 'X', 'Y', 'Z'\n};\n#define nl \"\\n\"\n#define ce cout<<\"\\n\"\n#define INF (1<<30)\n#define pb push_back\n#define vi vector<int>\n#define vl vector<lli>\n#define yes cout<<\"YES\"<<\"\\n\"\n#define no cout<<\"NO\"<<\"\\n\"\n#define S size()\n#define L length()\n#define B begin()\n#define E end()\n#define F first\n#define se second\n#define INF (1<<30)\n#define lcm(a,b) a*b/__gcd(a,b)\n#define for_t(i,j,n) for(int i=0,int j=n-1;i<j;i++,j--)\n#define g_arr(a,n) sort(arr,arr+n,greater<int>())\n#define dsort sort(vec.begin(), vec.end(), greater<int>())\n#define PI 3.1415926535897932384626433832795\n#define isort sort(vec.B,vec.E)\n#define fpre for(int i=1;i<n;i++){pre[i]=pre[i-1]+vec[i];}\n#define sp \" \"\n#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>\n#define fo(x,n) for (int i = x; i < n; ++i)\n#define fo_index(i,k,n) for ( int i = k; i <= n; ++i)\n#define rev_fo_index(i,k,n) for (int i = k; i >= n; --i)\n//--------------ordered set Functionality-----------------\n/*\n//-------------Ordered set declared with name o_set--------\n ordered_set o_set;\n \n//-------------Finding the second smallest element--------\n \n cout << *(o_set.find_by_order(1))<< endl;\n \n \n//------------Finding the number of elements strictly less than k=4--------------\n \n cout << o_set.order_of_key(4)<< endl;\n*/\n \n#ifndef ONLINE_JUDGE\n#define debug(x) cerr<< #x <<\"=\";_print(x);cerr<<nl;\n#else\n#define debug(x)\n#endif\nvoid _print(int a){cerr<<a;}\nvoid _print(lli a){cerr<<a;}\nvoid _print(double a){cerr<<a;}\nvoid _print(char a){cerr<<a;}\nvoid _print(bool a){cerr<<a;}\nvoid _print(string a){cerr<<a;}\n \ntemplate<class T>void _print(vector<T>vec){\n cerr <<\"[ \";\n for(T it:vec){\n cerr<<it<<\" , \";\n }\n cerr<<\" ]\";\n}\n template<class T>\nvoid _print(stack<T> s) {\n cerr << \"[ \";\n while (!s.empty()) {\n cerr << s.top() << \" , \";\n s.pop();\n }\n cerr << \"]\\n\";\n}\ntemplate<class T>void _print(set<T>st){\n cerr <<\"[ \";\n for(T it:st){\n cerr<<it<<\" , \";\n }\n cerr<<\" ]\";\n}\nvoid _print(const vector<vector<int>>& vec) {\n cerr << \"{\\n\";\n for (const auto& row : vec) { // Iterate through each row\n cerr << \" { \";\n for (const auto& element : row) { // Iterate through each element in the row\n cerr << element << \" \";\n }\n cerr << \"}\\n\";\n }\n cerr << \"}\\n\";\n}\nvoid _print(const set<vector<int>>& vecSet) {\n cerr << \"{\\n\"; // Start of the set\n for (const auto& vec : vecSet) { // Iterate through each vector in the set\n cerr << \" { \";\n for (const auto& element : vec) { // Iterate through each element in the vector\n cerr << element << \" \"; // Print each element\n }\n cerr << \"}\\n\"; // End of the vector\n }\n cerr << \"}\\n\"; // End of the set\n}\ntemplate<class T1, class T2>\nvoid _print(const vector<pair<T1, T2>>& vec) {\n cerr << \"{ \";\n for(const auto& pair : vec) {\n cerr << \"(\" << pair.first << \", \" << pair.second << \") \";\n }\n cerr << \"}\" << endl;\n}\n \ntemplate<class T1,class T2>\nvoid _print(const map<T1, T2>& mp) {\n cerr << \"{ \";\n for(const auto& pair : mp) {\n cerr <<\"( \" << pair.first << \" : \" << pair.second << \") \";;\n }\n cerr << \" }\" << nl;\n}\ntemplate<class T>\nvoid _print(const unordered_map<T, T>& mp) {\n cerr << \"{ \";\n for(const auto& pair : mp) {\n cerr <<\"( \" << pair.first << \" : \" << pair.second << \") \";;\n }\n cerr << \" }\" << nl;\n}\nvoid _print(priority_queue<int> pq) {\n cerr << \"{ \";\n // Temporary vector to restore the original state of priority_queue\n vector<int> temp;\n \n // Pop elements from priority_queue and print them\n while (!pq.empty()) {\n int top = pq.top();\n cerr << top << \" \";\n temp.push_back(top); // Store the element in temp vector\n pq.pop(); // Pop the element from priority_queue\n }\n \n // Restore the original state of priority_queue\n for (int num : temp) {\n pq.push(num);\n }\n \n cerr << \"}\" << endl;\n}\nvoid _print(priority_queue<pair<int, int>> pq) {\n cerr << \"{ \";\n \n // Temporary vector to restore the original state of priority_queue\n vector<pair<int, int>> temp;\n \n // Pop elements from priority_queue and print them\n while (!pq.empty()) {\n auto top = pq.top();\n cerr << \"(\" << top.first << \",\" << top.second << \") \";\n temp.push_back(top); // Store the element in temp vector\n pq.pop(); // Pop the element from priority_queue\n }\n \n // Restore the original state of priority_queue\n for (auto& elem : temp) {\n pq.push(elem);\n }\n \n cerr << \"}\" << endl;\n}\ntemplate<class T>\nvoid _print(const vector<pair<T, T>>& vec) {\n cerr << \"{ \";\n for(const auto& pair : vec) {\n cerr << \"(\" << pair.first << \", \" << pair.second << \") \";\n }\n cerr << \"}\" << endl;\n}\ntemplate<class T,class V>void _print(pair<T,V>pa){\n cerr <<\"[ \";\n _print(pa.F);\n cerr<<sp;\n _print(pa.se);\n cerr<<\" ]\";\n}\nconst int N=1e6+10;\nlli M=1e9+7;\nint lim=1e7+10;\n \nint arr[N]={0};\nint prime[N];\nint fact[N];\nint inver_fact[N];\n \n// 2D vector with size...............\n// vector<vector<int>> vec( row_size , vector<int> (col_size));\n \nlli mini=1e18;\ntemplate<typename T, typename U>\nstatic inline void amin(T &x, U &y){if (y < x)x = y;}\n \nlli maxi=-1e18;\ntemplate<typename T, typename U>\nstatic inline void amax(T &x, U &y) {if (x < y) x = y; }\nlli mod(lli x){\n return ((x%M+M)%M);\n}\n \nlli power_Bigmod(lli x, lli y, lli p)\n{\n lli res = 1;\n while (y > 0) {\n if (y % 2 == 1)\n res = mod(mod(res) *1ll* mod(x));\n // y = y/2\n y = y >> 1;\n x = mod(mod(x) *1ll* mod(x));\n }\n return res % p;\n}\n \nlli add(lli a, lli b){\n return mod(mod(a)+mod(b));\n}\nlli mul(lli a,lli b){\n return mod(mod(a)*1ll*mod(b));\n}\nlli sub(lli a, lli b){\n return mod(mod(a)+mod(b)+ M);\n}\nlli Div(lli a,lli b){\n return mod(mod(a)*mod(power_Bigmod(b,M-2,M)));\n}\nlong double eDis(long double x1,long double y1,long double x2,long double y2){\n return sqrtl((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));\n}\nvoid factorial(){\n fact[0]=1;\n \n for(int i=1;i<N;i++){\n // debug(fact[i-1])\n fact[i]=(fact[i-1]*1ll*i)%M;\n inver_fact[i]=power_Bigmod(fact[i],M-2,M);\n \n }\n}\n \n \nlli nCr(lli n,lli r){\n if(n<r) return 0;\n lli res=fact[n];\n res=(res*power_Bigmod(fact[r],M-2,M))%M;\n res=(res*power_Bigmod(fact[n-r],M-2,M))%M;\n return res;\n}\n \n \nlli amin(lli x,lli y){\n if(x>y) return y;\n else return x;\n}\nlli amax(lli x,lli y){\n if(x<y)return y;\n else return x;\n}\n \n//set bit in position index\nint set_bit(int n,int index){\n return (n| (1<<index));\n \n}\n//Flip bit in position index\nint Flip_bit(int n,int index){\n return (n^ (1<<index));\n \n}\n \n// check is it set bit in position index\nint check_bit(int n,int index){\n return (n & (1<<index));\n}\n \nbool primality_test(lli n){\n if (n <= 1) return false;\n if (n <= 3) return true;\n if (n % 2 == 0 || n % 3 == 0) return false;\n for(lli i=5;i*i<=n;i++){\n if (n % i == 0) return false;\n \n }\n \n return true;\n \n}\n \nvoid seive(){\n for(int i=1;i<=N;i++){\n prime[i]=i;\n }\n for(int i=2;i*i<=N;i++){\n if(prime[i]==i){\n for(int j=2*i;j<=N;j+=i){\n prime[j]=i;\n }\n }\n }\n \n}\n \n \nvoid seive_for_segment_sieve(vector<long long int>&prime,long long int high){\n bool ck[high + 1];\n memset(ck, true, sizeof(ck));\n ck[1] = false;\n ck[0] = false;\n for (int i = 2; (i * i) <= high; i++) {\n if (ck[i] == true) {\n for (int j = i * i; j <= sqrt(high); j = j + i) {\n ck[j] = false;\n }\n }\n }\n for (int i = 2; i * i <= high; i++) {\n if (ck[i] == true) {\n prime.push_back(i);\n }\n }\n// debug(prime)\n}\n// Segment sieve\n \nvoid segment_sieve(long long low,long long int high){\n if (low<2 and high>=2){\n low = 2;\n }\n bool prime[high - low + 1];\n \n memset(prime, true, sizeof(prime));\n \n vector<long long int> chprime;\n seive_for_segment_sieve(chprime, high);\n \n for (int i : chprime) {\n int lower = (low / i);\n \n if (lower <= 1) {\n lower = i + i;\n }\n else if (low % i) {\n lower = (lower * i) + i;\n }\n else {\n lower = (lower * i);\n }\n for (int j = lower; j <= high; j = j + i) {\n prime[j - low] = false;\n }\n }\n \n for (int i = low; i <= high; i++) {\n if (prime[i - low] == true) {\n cout << (i) << \" \";\n }\n }\n}\n//count number of coprime\n \nvoid phi_1_to_n(int n) {\n vector<pair<int,int>> phi(n + 1);\n for (int i = 0; i <= n; i++)\n { phi[i].first = i; \n phi[i].se=i;}\n \n for (int i = 2; i <= n; i++) {\n if (phi[i].first == i) {\n for (int j = i; j <= n; j += i)\n {\n \n phi[j].first -= phi[j].first / i;\n \n \n }\n }\n }\n for(auto it:phi){cout<<it.F<<sp<<it.se<<nl;}\n}\nlli binary(lli n){\n lli low=0,high=1e9;\n while (low < high) {\n lli mid = (low + high+1)>>1;\n if (mid*(mid+1)/2<= n) {\n low = mid;\n } else {\n high = mid -1;\n }\n }\n return low;\n}\n//Harmonic Series\n//Calculate Total number of divisor\nvector<int>total_divisor(lim,1);\nvoid divisor_count(){\n for(int i=2;i<lim;i++){\n for(int j=i;j<lim;j+=i){\n total_divisor[j]++;\n }\n }\n \n}\n \n map<int,int>mp;\nvoid prime_factor(lli n){\n for(lli j=2;j*j<=n;j++){\n if(n%j==0){\n lli cnt=0;\n while(n%j==0){\n cnt++;\n n/=j;\n }\n mp[j]=cnt;\n \n }\n }\n if(n>1){\n mp[n]=1;\n }\n }\n \ndouble eDis(double x1,double y1,double x2,double y2){\n return sqrtl((x1-x2)*(x1-x2)+(y1-y2)*(y1-y2));\n}\n \nlli power(lli base, lli exponent) {\n lli result = 1;\n //base = base % modulus;\n while (exponent > 0) {\n if (exponent % 2 == 1) {\n result = (result * base) ;\n }\n exponent = exponent >> 1;\n base = (base * base);\n }\n return result;\n}\n \nstring bin(lli x)\n{\n // Convert to binary using 64 bits for large numbers\n string binary = bitset<64>(x).to_string();\n \n // Find the position of the first '1'\n size_t firstOnePos = binary.find('1');\n \n // If there is no '1', the number is zero\n if (firstOnePos != string::npos) {\n // Return the substring from the first '1' to the end\n return binary.substr(firstOnePos);\n }\n \n // Return \"0\" for the case where the number is zero\n return \"0\";\n}\n \n \ninline void MODE() {\n \n#ifndef ONLINE_JUDGE\n \nfreopen(\"inputf.txt\", \"r\", stdin);\n \nfreopen(\"outputf.txt\", \"w\", stdout);\n \nfreopen(\"expected output_txt.txt\",\"w\",stderr);\n \n#endif // ONLINE_JUDGE\n}\nstruct Comparator {\n bool operator()(const pair<int, int>& a, const pair<int, int>& b) const { \n if (a.first == b.first) {\n return a.second < b.second; // Compare second values if first values are equal\n }\n return a.first < b.first; // Compare first values\n }\n};\n \nlli gcd(lli a, lli b) {\n if(a<b){\n swap(a,b);\n }\n \n if (b == 0)\n return a;\n return gcd(b, a % b);\n}\nint ceil(int a,int b)\n{\n return (a+b-1)/b;\n}\n \n \nbool compareFirst(const vector<int>& a, const vector<int>& b) {\n return a[0] < b[0];\n}\n \n \n \nint path(int x,int y,int m,int n,int (&dp)[105][105]){\n if(x==m-1 and y==n-1)return 1;\n \n int ans1=0,ans2=0;\n \n if(dp[x][y]!=-1)return dp[x][y];\n \n if(x<m-1)\n ans1= path(x+1,y,m,n,dp);\n \n if(y<n-1)\n ans2= path(x,y+1,m,n,dp);\n // cout<<dp[x][y+1]+dp[x+1][y]<<nl;\n \n return dp[x][y]=ans1+ans2;\n \n \n}\n//find a string first occarance and find a string last occarance in a stiring\n/*\n size_t first_digit=str.find_first_of(s);\n size_t last_digit=str.find_last_of(s);\n */\n \nbool check(lli mid,vector<lli>&vec, lli n){\n \n lli temp=0;\n \n \n for(int i=0;i<n;i+=2){\n if(vec[i]+mid<vec[i+1] or i==n-1){\n if(temp){\n return false;\n \n }\n temp++;\n i--;\n }\n }\n return true;\n}\nint binarySearch( vector<lli>vec, int target) {\n int left = 0;\n int right = vec.size() - 1;\n \n int result = -1; // To store the index of the target if found\n \n while (left <= right) {\n int mid = left + (right - left) / 2;\n \n if (vec[mid] == target) {\n return mid;\n }\n else if (vec[mid] < target) {\n left = mid + 1;\n } else {\n right = mid - 1; \n }\n }\n \n \n return -1;\n}\n \n \nint findRoots(double a, double b, double c) {\n \n \n double discriminant = b * b - 4 * a * c;\n \n if (discriminant > 0) {\n // Two distinct real roots\n double root1 = (-b + sqrt(discriminant)) / (2 * a);\n double root2 = (-b - sqrt(discriminant)) / (2 * a);\n return max(root1,root2);\n } else if (discriminant == 0) {\n // One real root (repeated)\n double root = -b / (2 * a);\n return root;\n } else {\n // Complex roots\n double realPart = -b / (2 * a);\n double imagPart = sqrt(-discriminant) / (2 * a);\n return realPart;\n }\n return 0;\n}\nlong long binToDec(const string& binary) {\n return stoll(binary, nullptr, 2); // Convert from base 2 (binary) to decimal\n}\n \n \n lli n,a,b,c,d,e,f,g;\n bool flag=false;\nint calculateSum(const vector<int>& seq) {\n int sum = 0;\n for (int num : seq) {\n sum += num;\n }\n return sum;\n}\n \nvector<int> differenceSequence(const vector<int>& seq) {\n vector<int> diff;\n for (int i = 1; i < seq.size(); i++) {\n diff.push_back(seq[i] - seq[i - 1]);\n }\n return diff;\n}\nvoid solve() {\n \n int n;\n cin>>n;\n lli c,d;\n cin>>c>>d;\n \n vl vec(n*n);\n \n fo(0,n*n)cin>>vec[i];\n \n sort(vec.B,vec.E);\n int x=vec[0];\n \n for(int i=0;i<n;i++){\n int t=x;\n \n for(int j=0;j<n;j++){\n debug(t)\n int v=binarySearch(vec,t);\n if(v!=-1){\n vec[v]=-1;\n sort(vec.B,vec.E);\n }\n else{\n no;return;\n }\n t+=c;\n \n }\n x+=d;\n \n }\n yes;\n \n \n \n \n}\n \nint32_t main(){\n \n fast;\n MODE();\n// seive();\n //factorial();\n \n int test;\n cin>>test;\n \n for(int i=1;i<=test;i++){\n solve();\n }\n \n \n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 306441954,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\nt=int(input())\nfor x in range(t):\n n,c,d = input().split(' ')\n n = int(n)\n c = int(c)\n d = int(d)\n inp = list(map(int, input().split(' ')))\n done=0\n lol=[0]*n*n\n lol[0]=min(inp)\n for i in range(0,n):\n if(i!=0):\n lol[i*n]=lol[(i-1)*n]+c\n for j in range(1,n):\n lol[i*n+j]=lol[i*n+j-1]+d\n for o in lol:\n if o not in inp or lol.count(o) != inp.count(o):\n print(\"NO\")\n done = 1\n break\n if(done==0):\n print(\"YES\")\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255698358,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\nfor _ in range(int(input())):\n n, c, d = map(int, input().split())\n l = list(map(int, input().split()))\n ln = []\n for i in range(n):\n for j in range(n):\n ln.append(min(l) + c*i + d*j)\n l.sort()\n ln.sort()\n if l == ln:\n print('YES')\n else:\n print('NO')\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 256967732,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\nimport java.util.*;\n \nimport javax.print.DocFlavor.READER;\n \npublic class Main \n{\n \n static void test(int n,int c,int d,long arr[])\n {\n int len=(int)Math.sqrt((int)arr.length);\n Arrays.sort(arr);\n long temp=arr[0];\n \n long dp[][]=new long[len][len];\n for (int i = 0; i<len; i++) \n {\n dp[i][0]=temp;\n for (int j = 1; j <len; j++) \n {\n dp[i][j]=dp[i][j-1]+d; \n }\n temp=temp+c;\n }\n \n for (int i = 0; i<len; i++) \n {\n \n for (int j = 0; j <len; j++) \n {\n boolean flag=false;\n \n \n for (int k = 0; k < arr.length; k++) \n {\n if(arr[k]==dp[i][j])\n {\n arr[k]=-1;\n flag=true;\n break;\n \n } \n }\n if (!flag) \n {\n System.out.println(\"NO\");\n return; \n }\n \n \n \n }\n \n }\n System.out.println(\"YES\");\n \n \n \n \n \n }\n \n public static void main(String[] args) \n {\n \n Scanner sc=new Scanner(System.in);\n int n=sc.nextInt();\n \n while (n-->0) \n {\n int x=sc.nextInt();\n int a=sc.nextInt();\n int b=sc.nextInt();\n long arr[]=new long[x*x];\n for (int i = 0; i < arr.length; i++) \n {\n arr[i]=sc.nextLong(); \n }\n test(x, a, b,arr);\n \n \n }\n \n \n \n}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255679449,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\nimport java.util.*;\npublic class progressiveSquare{\n public static boolean check(int[] arr , int x ){\n for(int i=0;i<arr.length;i++){\n if(arr[i]==x){\n arr[i]=-arr[i];\n return true;\n }\n }\n return false;\n }\n public static void main(String[] args){\n Scanner input = new Scanner(System.in);\n int t = input.nextInt();\n while(t-->0){\n int n = input.nextInt();\n int c = input.nextInt();\n int d = input.nextInt();\n ArrayList<Integer> found = new ArrayList<>();\n \n int arr[] = new int[n*n];\n int row[] = new int[n];\n int col[] = new int[n];\n int a=Integer.MAX_VALUE;\n for(int i=0;i<n*n;i++){\n arr[i]=input.nextInt();\n a = Math.min(a, arr[i]);\n }\n int flag=1;\n for(int i = 0 ; i<n ; i++){\n col[i] = a + c*i;\n // System.out.println(col[i]+ \" col\");\n if(!check(arr,col[i])){\n flag=0;\n break;\n }\n else{\n found.add(col[i]);\n }\n }\n int x;\n for(int i =0; i<n;i++){\n for(int j = 0 ; j<n-1; j++){\n x = col[i] + d*(j+1);\n // System.out.println(x+ \" x\");\n if(!check(arr,x)){\n flag=0;\n break;\n }\n else{\n found.add(x);\n }\n }\n }\n // System.out.println(found);\n if(flag==1 && found.size() == n*n){\n System.out.println(\"yes\");\n }\n else{\n System.out.println(\"No\");\n }\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255705958,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Progressive Square\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: A progressive square of size $$$n$$$ is an $$$n \\times n$$$ matrix. Maxim chooses three integers $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$ and constructs a progressive square according to the following rules:\n\n$$$$$$a_{i+1,j} = a_{i,j} + c$$$$$$\n\n$$$$$$a_{i,j+1} = a_{i,j} + d$$$$$$\n\nFor example, if $$$n = 3$$$, $$$a_{1,1} = 1$$$, $$$c=2$$$, and $$$d=3$$$, then the progressive square looks as follows:\n\n$$$$$$ \\begin{pmatrix} 1 & 4 & 7 \\\\ 3 & 6 & 9 \\\\ 5 & 8 & 11 \\end{pmatrix} $$$$$$\n\nLast month Maxim constructed a progressive square and remembered the values of $$$n$$$, $$$c$$$, and $$$d$$$. Recently, he found an array $$$b$$$ of $$$n^2$$$ integers in random order and wants to make sure that these elements are the elements of that specific square.\n\nIt can be shown that for any values of $$$n$$$, $$$a_{1,1}$$$, $$$c$$$, and $$$d$$$, there exists exactly one progressive square that satisfies all the rules.\n\nInput: The first line contains an integer $$$t$$$ ($$$1 \\le t \\le {10} ^ 4$$$)ย โ the number of test cases.\n\nThe first line of each test case contains three integers $$$n$$$, $$$c$$$, and $$$d$$$ ($$$2 \\le n \\le 500$$$, $$$1 \\le c, d \\le 10^6$$$)ย โ the size of the square and the values of $$$c$$$ and $$$d$$$ as described in the statement.\n\nThe second line of each test case contains $$$n \\cdot n$$$ integers $$$b_1, b_2, \\dots, b_{n \\cdot n}$$$ ($$$1 \\le b_i \\le 10^9$$$)ย โ the elements found by Maxim.\n\nIt is guaranteed that the sum of $$$n ^ 2$$$ over all test cases does not exceed $$$25 \\cdot {10} ^ 4$$$.\n\nOutput: For each test case, output \"YES\" in a separate line if a progressive square for the given $$$n$$$, $$$c$$$, and $$$d$$$ can be constructed from the array elements $$$a$$$, otherwise output \"NO\".\n\nYou can output each letter in any case (lowercase or uppercase). For example, the strings \"yEs\", \"yes\", \"Yes\", and \"YES\" will be accepted as a positive answer.\n\nExamples:\ninput:\n5\n3 2 3\n3 9 6 5 7 1 10 4 8\n3 2 3\n3 9 6 5 7 1 11 4 8\n2 100 100\n400 300 400 500\n3 2 3\n3 9 6 6 5 1 11 4 8\n4 4 4\n15 27 7 19 23 23 11 15 7 3 19 23 11 15 11 15\noutput:\nNO\nYES\nYES\nNO\nNO\n\nNote: \n\nBuggy Code:\n// clang-format off\n \n#ifdef ONLINE_JUDGE\n#pragma GCC optimize(\"Ofast\")\n#pragma GCC optimize(\"unroll-loops\")\n#define dbg(...)\n#define SOLVE solve();\n#endif\n \n#include <bits/stdc++.h>\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\n \nconstexpr int mod = 998244353; // 1e9 + 7\n \nusing namespace std;\nusing namespace __gnu_pbds;\ntemplate <typename T>\nusing ost = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;\n \nmt19937 rng(chrono::steady_clock::now().time_since_epoch().count());\n#define uid(a, b) uniform_int_distribution<int>(a, b)(rng)\n \ntemplate <const int T> struct ModInt { const static int mod = T; int x; ModInt(int x = 0) : x(x % mod) {} ModInt(long long x) : x(int(x % mod)) {} int val() { return x; } ModInt operator+(const ModInt &a) const { int x0 = x + a.x; return ModInt(x0 < mod ? x0 : x0 - mod); } ModInt operator-(const ModInt &a) const { int x0 = x - a.x; return ModInt(x0 < 0 ? x0 + mod : x0); } ModInt operator*(const ModInt &a) const { return ModInt(1ll * x * a.x % mod); } ModInt operator/(const ModInt &a) const { return *this * a.inv(); } void operator+=(const ModInt &a) { x += a.x; if (x >= mod) x -= mod; } void operator-=(const ModInt &a) { x -= a.x; if (x < 0) x += mod; } void operator*=(const ModInt &a) { x = 1LL * x * a.x % mod; } void operator/=(const ModInt &a) { *this = *this / a; } ModInt pow(int64_t n) const { ModInt res(1), mul(x); while (n) { if (n & 1) res *= mul; mul *= mul; n >>= 1; } return res; } ModInt inv() const { int a = x, b = mod, u = 1, v = 0; while (b) { int t = a / b; a -= t * b; swap(a, b); u -= t * v; swap(u, v); } if (u < 0) u += mod; return u; } };\n \ntypedef ModInt<mod> mint; typedef long long ll; typedef vector<int> vi; typedef vector<vector<int>> vvi; typedef vector<long long> vl; typedef vector<vector<long long>> vvl; typedef vector<pair<int, int>> vpi; typedef vector<pair<long long, long long>> vpl; typedef pair<int, int> pi; typedef pair<long long, long long> pl; typedef unsigned int uint; typedef unsigned long long ull;\n \ntemplate <typename T> using vc = vector<T>; template <typename T> using pq = priority_queue<T>; template <typename T> using qu = queue<T>; template <typename T> using dq = deque<T>; template <typename T> using stk = stack<T>; template <typename T1, typename T2> using pr = pair<T1, T2>; template <typename T> using mst = multiset<T>; template <typename Key, typename T> using ump = unordered_map<Key, T>; template <typename T> using ust = unordered_set<T>; template <typename T> using mpq = priority_queue<T, vector<T>, greater<T>>;\n \nvoid __print(int x) { cerr << x; } void __print(long x) { cerr << x; } void __print(long long x) { cerr << x; } void __print(unsigned x) { cerr << x; } void __print(unsigned long x) { cerr << x; } void __print(unsigned long long x) { cerr << x; } void __print(float x) { cerr << x; } void __print(double x) { cerr << x; } void __print(long double x) { cerr << x; } void __print(char x) { cerr << '\\'' << x << '\\''; } void __print(const char *x) { cerr << '\\\"' << x << '\\\"'; } void __print(const string x) { cerr << '\\\"' << x << '\\\"'; } void __print(bool x) { cerr << (x ? \"true\" : \"false\"); } void __print(mint a) { cerr << a.x; } template <typename T, typename V> void __print(pair<T, V> x) { cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}'; } template <typename T> void __print(T x) { int f = 0; cerr << '{'; for (auto &i : x) cerr << (f++ ? \",\" : \"\"), __print(i); cerr << \"}\"; } template <typename T> void __print(stack<T> x) { vector<T> _x; while (!x.empty()) _x.push_back(x.top()), x.pop(); reverse(_x.begin(), _x.end()); __print(_x); } template <typename T> void __print(queue<T> x) { vector<T> _x; while (!x.empty()) _x.push_back(x.front()), x.pop(); reverse(_x.begin(), _x.end()); __print(_x); } template <typename T> void __print(priority_queue<T> x) { vector<T> _x; while (!x.empty()) _x.push_back(x.top()), x.pop(); reverse(_x.begin(), _x.end()); __print(_x); } template <typename T> void __print(priority_queue<T, vector<T>, greater<T>> x) { vector<T> _x; while (!x.empty()) _x.push_back(x.top()), x.pop(); reverse(_x.begin(), _x.end()); __print(_x); } void _print() { cerr << \"]\\n\\n\"; } template <typename T, typename... V> void _print(T t, V... v) { __print(t); if (sizeof...(v)) cerr << \", \"; _print(v...); }\n#ifndef ONLINE_JUDGE\n#define dbg(x...) cerr << \"[\" << #x << \"] = [\"; _print(x)\n#define SOLVE cerr << (_ > 0 ? \"\\n####### TestCase \" : \"####### TestCase \") << _ + 1 << \" #######\\n\\n\"\n#endif\n \n#define reif(t, ...) if (t) { __VA_ARGS__; return; }\n#define coif(t, ...) if (t) { __VA_ARGS__; continue; }\n#define brif(t, ...) if (t) { __VA_ARGS__; break; }\n \n#define all(x) x.begin(), x.end()\n#define rev(x) reverse(all(x))\n#define s0rt(x) sort(all(x))\n#define s0rtd(x) sort(all(x)), rev(x);\n#define s0rtx(x, ...) sort(all(x), __VA_ARGS__);\n#define sum(x) accumulate(all(x), 0ll)\n#define len(x) (long long)(x.size())\n#define mine(v) *min_element(all(v))\n#define maxe(v) *max_element(all(v))\n#define unq(x) s0rt(x), x.erase(unique(all(x)), x.end()), x.shrink_to_fit()\n#define FOR1(a) for (int _ = 0; _ < (a); ++_)\n#define FOR2(i, a) for (int i = 0; i < (a); ++i)\n#define FOR3(i, a, b) for (int i = a; i < (b); ++i)\n#define FOR4(i, a, b, c) for (int i = a; i < (b); i += (c))\n#define FOR5(i, a, condn, b, add) for (int i = a; condn < b; i += add)\n#define FOR1_R(a) for (int _ = (a); _ >= (0); --i)\n#define FOR2_R(i, a) for (int i = (a); i >= (0); --i)\n#define FOR3_R(i, a, b) for (int i = (a); i >= (b); --i)\n#define FOR4_R(i, a, b, c) for (int i = (a); i >= (b); i -= c)\n#define FOR5_R(i, a, condn, b, add) for (int i = (a); condn >= (b); i -= add)\n#define TRAV1(i, a) for (auto &i : a)\n#define TRAV2(i, j, a) for (auto &[i, j] : a)\n#define LB1(x, y) lower_bound(all(x), y)\n#define UB1(x, y) upper_bound(all(x), y)\n#define LB2(x) lower_bound(x)\n#define UB2(x) upper_bound(x)\n#define CNT1(x, y) count(all(x), y)\n#define cnt2(x) count(x)\n#define overload5(a, b, c, d, e, f, ...) f\n#define overload2(a, b, c, d, ...) d\n#define overload1(a, b, c, ...) c\n#define f0r(...) overload5(__VA_ARGS__, FOR5, FOR4, FOR3, FOR2, FOR1)(__VA_ARGS__)\n#define f0rd(...) overload5(__VA_ARGS__, FOR5_R, FOR4_R, FOR3_R, FOR2_R, FOR1_R)(__VA_ARGS__)\n#define trav(...) overload2(__VA_ARGS__, TRAV2, TRAV1)(__VA_ARGS__)\n#define lb(...) overload1(__VA_ARGS__, LB1, LB2)(__VA_ARGS__)\n#define ub(...) overload1(__VA_ARGS__, UB1, UB2)(__VA_ARGS__)\n#define c0unt(...) overload1(__VA_ARGS__, CNT1, CNT2)(__VA_ARGS__)\n#define pb push_back\n#define eb emplace_back\n#define ep emplace\n#define ins insert\n#define ers erase\n#define beg begin()\n#define ed end()\n#define back back()\n#define mkp make_pair\n#define f first\n#define s second\n#define yes \"YES\"\n#define no \"NO\"\n#define fun(type, name, ...) auto name = [&](__VA_ARGS__) -> type\n#define recfun(type, name, ...) auto name = [&](auto &&name, __VA_ARGS__) -> type\n \nint popcnt(int x) { return __builtin_popcount(x); } int topbit(int x) { return (x == 0 ? -1 : 31 - __builtin_clz(x)); } int lowbit(int x) { return (x == 0 ? -1 : __builtin_ctz(x)); } int popcnt(long long x) { return __builtin_popcountll(x); } int topbit(long long x) { return (x == 0 ? -1 : 63 - __builtin_clzll(x)); } int lowbit(long long x) { return (x == 0 ? -1 : __builtin_ctzll(x)); }\n \ntemplate <typename T> int sgn(const T &x) { return (x > 0 ? 1 : x < 0 ? -1 : 0); }; template <typename T, typename U> T ceil(T x, U y) { return (x > 0 ? (x + y - 1) / y : x / y); } template <typename T, typename U> T floor(T x, U y) { return (x > 0 ? x / y : (x - y + 1) / y); } template <class T, class S> bool chmx(T &a, const S &b) { return (a < b ? a = b, 1 : 0); } template <class T, class S> bool chmn(T &a, const S &b) { return (a > b ? a = b, 1 : 0); }\n \ntemplate <typename U, typename T> T binser(U check, T _lo, T _hi, bool _minimize) { T _ret = (_minimize ? _hi + 1 : _lo - 1); while (_lo <= _hi) { T _mid = _lo + _hi >> 1ll; if (_minimize) { if (check(_mid)) { _hi = _mid - 1; _ret = _mid; } else { _lo = _mid + 1; } } else { if (check(_mid)) { _lo = _mid + 1; _ret = _mid; } else { _hi = _mid - 1; } } } return _ret; }\ntemplate <typename T> void prfxsum(vector<T> &a) { int _n = a.size(); for (int i = 1; i < _n; i++) a[i] += a[i - 1]; };\ntemplate <typename T, typename U> vector<T> getorder(T start, T size, U comp) { vector<T> ord(size); iota(ord.begin(), ord.end(), start); sort(ord.begin(), ord.end(), comp); return ord; }\ntemplate <typename T = int, typename U = int> map<T, U> mappos(const vector<T> &data, U offset) { map<T, U> ps; for (int i = 0; i < data.size(); i++) ps[data[i]] = i + offset; return ps; }\ntemplate <typename T> map<T, int> counter(const vector<T> &data) { map<T, int> x; for (auto &i : data) x[i]++; return x; };\ntemplate <typename T, typename U> T toset(const vector<U> &data) { T x; for (auto &i : data) x.insert(i); return x; };\n \ntemplate <typename T, typename U> U safeErase(T &setik, U target) { auto iterator = setik.find(target); assert(iterator != setik.end()); U a = *iterator; setik.erase(iterator); return a; }\ntemplate <typename T, typename U> pair<U, bool> Erase(T &setik, U target) { auto iterator = setik.find(target); pair<U, bool> a = {0, false}; if (iterator != setik.end()) { a.first = *iterator; setik.erase(iterator); } return a; }\n \ntemplate <typename T> T pop(multiset<T> &que, bool front = 1) { assert(que.empty() == false); auto iterator = front ? que.begin() : prev(que.end()); T a = *iterator; que.erase(iterator); return a; } template <typename T> T pop(set<T> &que, bool front = 1) { assert(que.empty() == false); auto iterator = front ? que.begin() : prev(que.end()); T a = *iterator; que.erase(iterator); return a; } template <typename T> T pop(stack<T> &que) { T a = que.top(); que.pop(); return a; } template <typename T> T pop(queue<T> &que) { T a = que.front(); que.pop(); return a; } template <typename T> T pop(deque<T> &que, bool front) { T a = (front ? que.front() : que.back()); (front ? que.pop_front() : que.pop_back()); return a; } template <typename T> T pop(priority_queue<T> &que) { T a = que.top(); que.pop(); return a; } template <typename T> T pop(priority_queue<T, vector<T>, greater<T>> &que) { T a = que.top(); que.pop(); return a; } template <typename T> T pop(vector<T> &que) { T a = que.back; que.pop_back(); return a; }\n \n// condition args(i, stack.top())\ntemplate <typename T, typename U> vector<int> nearestelem(const vector<T> &a, U condition, bool reverse = false, int def = -1) { int n = a.size(); stack<int> s; vector<int> ans(n, def); auto go = [&](int i) { while (!s.empty() && condition(i, s.top())) { ans[s.top()] = i; s.pop(); } s.push(i); }; if (reverse) for (int i = n - 1; i > -1; i--) go(i); else for (int i = 0; i < n; i++) go(i); return ans; }\ntemplate <typename T, typename U> constexpr T fpow(T x, U n) { T result = 1; while (n > 0) { if (n & 1ll) result *= x; x *= x; n >>= 1ll; } return result; }\ntemplate <typename T> struct prfx { vector<T> _a; int _n; prfx(const vector<T> &a) { _n = a.size() + 1; _a.resize(_n); for (int i = 0; i < _n - 1; i++) _a[i + 1] = _a[i] + a[i]; } T get(int x, int y) { if (y < x) return 0; return _a[y + 1] - _a[x]; } };\nvector<int> divisors(int V) { vector<int> ret; for (int i = 1; (long long)i * i < V + 1; ++i) { if (V % i == 0) { ret.push_back(i); if (V / i != i) { ret.push_back(V / i); } } } return ret; }\n \nconst int imx = INT_MAX;\nconst int imn = INT_MIN;\nconst int lmx = LLONG_MAX;\nconst int lmn = LLONG_MIN;\n \n// clang-format on\n \nconst int N = 3e5;\n \nll n, c, d;\nll a[N], b[N];\n \nbool verify(int x) {\n int ix = 0;\n \n f0r(i, 0, n) {\n b[ix++] = x;\n f0r(j, 1, n) b[ix++] = x + (ll)j * d;\n x += c;\n }\n \n sort(b, b + n * n);\n \n f0r(i, n * n) if (a[i] != b[i]) return false;\n return true;\n}\n \nvoid solve() {\n cin >> n >> c >> d;\n f0r(i, n * n) cin >> a[i];\n \n sort(a, a + n * n);\n \n f0r(i, n * n) reif(verify(a[i]), cout << \"YES\" << '\\n');\n \n cout << \"NO\" << '\\n';\n}\n \nint main() {\n cin.tie(0)->ios_base::sync_with_stdio(false);\n \n int testcases = 0;\n cin >> testcases;\n \n f0r(testcases) solve();\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1955B",
"submission_id": 255670090,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\n// Awwawa! Dis cold yis ratten buy kitten!\n#include <bits/stdc++.h>\n#define all(x) (x).begin(), (x).end()\n \ntemplate <typename T>\nstd::ostream& operator<<(std::ostream& os, const std::vector<T>& v) {\n int n = v.size();\n for (int i = 0; i < n; ++i)\n os << v[i] << ' ';\n return os;\n}\nusing namespace std;\nusing i64 = long long;\n \nint main() {\n ios::sync_with_stdio(false);\n cin.tie(nullptr);\n int T;\n cin >> T;\n while (T--) {\n int n, q;\n cin >> n >> q;\n vector<int> a(n + 1), occur(n + 1);\n for (int i = 1; i <= n; ++i) {\n cin >> a[i];\n occur[a[i]] += 1;\n }\n \n sort(all(a));\n int flag = (n % 2 == 0 ? (n + 2) / 2 : (n + 1) / 2);\n while (q--) {\n int x;\n cin >> x;\n int cnt = a.end() - lower_bound(all(a), x);\n // cerr << cnt << '\\n';\n int ans = -1;\n if (cnt < flag) {\n ans = a[flag - cnt];\n } else {\n // int beg = lower_bound(all(a), x) - a.begin();\n // cerr << beg << '\\n';\n int f = flag;\n for (int i = 0; i < x; ++i) {\n int cc = occur[i];\n for (int j = i; j <= n; j += x) {\n cc += occur[j];\n }\n if (f <= cc) {\n ans = i;\n break;\n }\n f -= cc;\n }\n \n // ans = a[flag] % x;\n }\n assert(ans != -1);\n cout << ans << ' ';\n }\n cout << '\\n';\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 279221950,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\n/*\n JAI SHREE RAM\n*/\n \n#include<bits/stdc++.h>\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\n#define int long long\n#define ld long double\n#define nl cout<<endl\n#define notpos cout<<-1<<endl\n#define sqrt sqrtl\n#define yes cout<<\"YES\"<<endl\n#define no cout<<\"NO\"<<endl\n#define all(v) v.begin(),v.end()\n#define f(i,n) for(int i=0;i<n;i++)\n#define vii vector<vector<int>>\n#define pii pair<int,int>\n#define inf LLONG_MAX\n#define pb push_back\n#define ff first\n#define ss second\n#define gsum(a) accumulate(a.begin(),a.end(),0LL)\n#define gmax(a) *max_element(a.begin(),a.end())\n#define gmin(a) *min_element(a.begin(),a.end())\n#define d1(x) cout<<#x<<\" \"<<x<<endl;\n#define d2(x,y) cout<<#x<<\" \"<<x<<\" \"<<#y<<\" \"<<y<<endl;\n#define d3(x,y,z) cout<<#x<<\" \"<<x<<\" \"<<#y<<\" \"<<y<<\" \"<<#z<<\" \"<<z<<endl;\n \nusing namespace __gnu_pbds;\nusing namespace std;\n \ntypedef pair<int,int> node;\ntypedef tree<node, null_type, less<node>,\n rb_tree_tag, tree_order_statistics_node_update> ordered_set;\n \nconst int mod = 1e9 + 7;\n//const int mod = 998244353;\nconst ld eps = 1e-7;\n \nint lcm(int a,int b){\n return (a/(__gcd(a,b)))*b;\n}\n \nvoid out(vector<int> v){\n for(auto x : v){\n cout<<x<<\" \";\n }\n nl;\n}\n \nvoid fastio(){\n std::ios::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n}\n \nint power(int a, int b, int p)\n {\n if(a==0)\n return 0;\n int res=1;\n a%=p;\n while(b>0)\n {\n if(b&1)\n res=(1ll*res*a)%p;\n b>>=1;\n a=(1ll*a*a)%p;\n }\n return res;\n }\n \nint N = 2e5 + 5;\nvector<int> spf(N,0);\nvector<int> path;\n \nvoid sieve(){\n spf[0] = 1;\n spf[1] = 1;\n for(int i=2;i<N;i++) spf[i] = i;\n \n for(int i=2;i*i<N;i++){\n if(spf[i] == i){\n for(int j=i*i;j<N;j+=i){\n if(spf[j] == j) spf[j] = i;\n }\n }\n }\n}\n \nvector<int> fact(N);\nvector<int> inv(N);\n \nvoid precompute(){\n fact[1] = 1;\n for(int i=2;i<N;i++){\n fact[i] = fact[i-1]*i;\n fact[i] %= mod;\n }\n inv[0] = 1;\n inv[1] = 1;\n for(int i=2;i<N;i++){\n inv[i] = power(fact[i],mod-2,mod);\n inv[i] %= mod;\n }\n}\n \nint nCr(int n,int r){\n int res = ((fact[n]*inv[r])%mod);\n res *= inv[n-r];\n res %= mod;\n return res;\n}\n \nint32_t main(){\n fastio();\n int tt=1;\n cin>>tt;\n while(tt--){\n int n,q;\n cin>>n>>q;\n vector<int> v(n);\n f(i,n) cin>>v[i];\n \n sort(all(v));\n vector<int> pref(n,0);\n pref[0]=v[0];\n for(int i=1;i<n;i++) pref[i] = pref[i-1]+v[i];\n \n vector<vector<int>> g(317);\n for(int i=1;i<317;i++){\n f(j,n){\n g[i].pb((v[j]%i));\n }\n sort(all(g[i]));\n }\n \n int m;\n if(n%2) m = (n+1)/2;\n else m = (n+2)/2;\n vector<int> ans;\n \n while(q--){\n int x;\n cin>>x;\n if(x<317) ans.pb(g[x][m-1]);\n else{\n int lo=0,hi=x-1,mid;\n int res = x-1;\n while(lo<=hi){\n mid = (lo+hi)/2;\n int mn = min(mid,x-1);\n int s = upper_bound(all(v),mn) - v.begin();\n int tot=0;\n \n for(int i=1;i*x<=v.back();i++){\n int l=i*x;\n int r=i*x + mid;\n \n int it1 = upper_bound(all(v),r) - v.begin();\n int it2 = lower_bound(all(v),l) - v.begin();\n \n tot += (it2-it1);\n if(i*x > v.back()) break;\n }\n if(s+tot >= m){\n res=mid;\n hi=mid-1;\n }\n else lo = mid+1;\n }\n ans.pb(res);\n }\n }\n out(ans);\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 279170772,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\n/**\n * https://codeforces.com/contest/2008/submission/280763922\n *\n * (c) 2024 Diego Sogari\n */\n#include <bits/stdc++.h>\n \nusing namespace std;\n \n#ifdef ONLINE_JUDGE\n#define debug\n#else\n#include \"debug.h\"\ninit();\n#endif\n \ntemplate <typename T> ostream &operator<<(ostream &os, const vector<T> &a) {\n return ranges::for_each(a, [&os](auto &ai) { os << ai << ','; }), os;\n}\nvoid println1(auto &&...args) { ((cout << args << ';'), ...) << endl; }\nvoid println(auto &&...args) { ((cout << args << ' '), ...) << endl; }\n \ntemplate <typename T> struct Num {\n T x;\n Num() { cin >> x; }\n Num(T a) : x(a) {}\n operator T &() { return x; }\n operator T() const { return x; }\n};\nusing Int = Num<int>;\n \nstruct Iota : vector<int> {\n Iota(int n, int s = 0) : vector<int>(n) { iota(begin(), end(), s); }\n Iota(int n, auto &&f, int s = 0) : Iota(n, s) { ranges::sort(*this, f); }\n};\n \nvoid solve(int t) {\n Int n, m;\n vector<Int> a(n), q(m);\n if (t == 42) {\n println1(n, m, a, q);\n return;\n }\n vector<int> cnt(n), ans(m);\n auto f = [&](int i, int j) { return q[i] < q[j]; };\n Iota idx(m, f);\n ranges::sort(a);\n for (int i = 0, j = 0; i < m; i++) {\n auto x = q[idx[i]];\n for (; a[j] < x; j++) {\n cnt[a[j]]++;\n }\n for (int k = j; k < n; k++) {\n cnt[a[k] % x]++;\n }\n for (int k = 0, c = 0; k < n; k++) {\n c += cnt[k];\n if (c > n / 2) {\n ans[idx[i]] = k;\n break;\n }\n }\n for (int k = j; k < n; k++) {\n cnt[a[k] % x]--;\n }\n }\n for (auto &&ai : ans) {\n cout << ai << ' ';\n }\n cout << endl;\n // println(ans);\n}\n \nint main() {\n cin.tie(nullptr)->tie(nullptr)->sync_with_stdio(false);\n Int t;\n for (int i = 1; i <= t; i++) {\n solve(i);\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 280766704,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\ndef check(mid,tar,q):\n cnt=0\n for i in range(0,n+1,q):\n if i==0:\n cnt+=qzh[mid]\n else:\n if i+mid>n:\n cnt+=qzh[n]-qzh[i-1]\n continue\n cnt+=qzh[i+mid]-qzh[i-1]\n \n if cnt<tar:\n return False\n else:\n return True\nfor i in range(int(input())):\n n,q=map(int,input().split())\n ns=list(map(int,input().split()))\n qzh=[0]*(n+1)\n tar=(n+1)//2 if n%2==1 else (n+2)//2\n als=[]\n for i in ns:\n qzh[i]+=1 \n for i in range(1,n+1):\n qzh[i]+=qzh[i-1]\n res={}\n for i in range(q):\n dis=int(input())\n if dis in res:\n print(res[dis])\n continue \n l=0\n r=dis-1\n while l<r:\n mid=(l+r)//2\n if check(mid,tar,dis):\n r=mid \n else:\n l=mid+1 \n res[dis]=l\n als.append(res[dis])\n print(*als)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 279280590,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\ndef sakurako_test(t, test_cases):\n results = []\n \n for _ in range(t):\n n, q = test_cases[_]['n_q']\n a = test_cases[_]['a']\n queries = test_cases[_]['queries']\n \n # Precompute the remainders for each x from 1 to n\n freq = [{} for _ in range(n + 1)]\n for x in range(1, n + 1):\n for ai in a:\n rem = ai % x\n if rem in freq[x]:\n freq[x][rem] += 1\n else:\n freq[x][rem] = 1\n \n # Function to find the median\n def find_median(freq_x):\n sorted_freq = sorted(freq_x.items())\n total = sum(freq_x.values())\n mid = (total // 2) + 1\n count = 0\n for val, cnt in sorted_freq:\n count += cnt\n if count >= mid:\n return val\n \n # Answer the queries\n answers = []\n for x in queries:\n median = find_median(freq[x])\n answers.append(median)\n \n results.append(answers)\n \n return results\n \n# Example usage\nif __name__ == \"__main__\":\n t = 2\n test_cases = [\n {'n_q': (5, 5), 'a': [1, 2, 3, 4, 5], 'queries': [1, 2, 3, 4, 5]},\n {'n_q': (6, 3), 'a': [1, 2, 6, 4, 1, 3], 'queries': [2, 1, 5]}\n ]\n \n results = sakurako_test(t, test_cases)\n for result in results:\n print(' '.join(map(str, result)))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 294857452,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\nt = int(input())\n \nfor _ in range(t):\n n, q = list(map(int, input().split()))\n arr = list(map(int, input().split()))\n ans = []\n for _ in range(q):\n k = int(input())\n if k == 1: \n ans.append(k-1)\n continue\n cs = [0]*k\n r = (n+1)//2 if n%2 else (n+2)//2\n for a in arr:\n cs[a % k] += 1\n for i in range(1, k):\n cs[i] += cs[i-1]\n if cs[i] >= r:\n ans.append(i)\n break \n print(*ans)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 279226667,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\nboolean MULTI_CASE = true;\nboolean ALWAYS_FLUSH = false;\n \nvoid go() {\n int n = nextInt();\n int qn = nextInt();\n int[] a = new int[n];\n for (int i = 0; i < a.length; ++i) a[i] = nextInt();\n int[] q = new int[qn];\n for (int i = 0; i < q.length; ++i) q[i] = nextInt();\n int[] g = new int[n + 1];\n for (int x : a) g[x]++;\n c = new int[n + 1];\n for (int i = 0; i < c.length; i++) c[i] = g[i] + (i == 0 ? 0 : c[i - 1]);\n int[] ret = new int[n + 1];\n for (int i = 2; i <= n; i++) {\n int l = 0, r = i - 1;\n if (check(l, i, n / 2 + 1)) ret[i] = l;\n else {\n while (l + 1 < r) {\n int mid = l + (r - l) / 2;\n if (check(mid, i, n / 2 + 1)) r = mid;\n else l = mid;\n }\n }\n ret[i] = r;\n }\n for (int x : q) {\n so(ret[x] + \" \");\n }\n sl();\n}\n \nint[] c;\n \nboolean check(int mid, int m, int reqCnt) {\n int cnt = 0;\n for (int i = 0; i < c.length; i += m) {\n cnt += c[Math.min(i + mid, c.length - 1)] - (i == 0 ? 0 : c[i - 1]);\n }\n return cnt >= reqCnt;\n}\n \n/**\n * Input.\n */\nInputStream inStream;\nbyte[] inBuff = new byte[1024];\nint inBuffCursor = 0, inBuffLen = 0;\n \nboolean isVisibleChar(int c) {\n return 33 <= c && c <= 126;\n}\n \nboolean hasNextByte() {\n if (inBuffCursor < inBuffLen) return true;\n inBuffCursor = 0;\n try {\n inBuffLen = inStream.read(inBuff);\n } catch (Exception e) {\n e.printStackTrace(System.out);\n }\n return inBuffLen > 0;\n}\n \nboolean hasNext() {\n while (hasNextByte() && !isVisibleChar(inBuff[inBuffCursor])) inBuffCursor++;\n return hasNextByte();\n}\n \nint nextByte() {\n return hasNextByte() ? inBuff[inBuffCursor++] : -1;\n}\n \nString next() {\n if (!hasNext()) throw new RuntimeException(\"no next.\");\n StringBuilder sb = new StringBuilder();\n int b = nextByte();\n while (isVisibleChar(b)) {\n sb.appendCodePoint(b);\n b = nextByte();\n }\n return sb.toString();\n}\n \nlong nextLong() {\n if (!hasNext()) throw new RuntimeException(\"no next.\");\n long result = 0;\n boolean negative = false;\n int b = nextByte();\n if (b < '0') {\n if (b == '-') negative = true;\n else if (b != '+') throw new RuntimeException(\"long number must start with +/-.\");\n b = nextByte();\n }\n while (isVisibleChar(b)) {\n if (b < '0' || b > '9') throw new RuntimeException(\"wrong digit in long:\" + (char) b);\n // NOTE: may overflow here, even if it is a valid Long, eg.-(1<<63).\n result = result * 10 + (b - '0');\n b = nextByte();\n }\n return negative ? -result : result;\n}\n \nint nextInt() {\n long x = nextLong();\n if (x < Integer.MIN_VALUE || x > Integer.MAX_VALUE)\n throw new RuntimeException(\"int overflow:\" + x);\n return (int) x;\n}\n \ndouble nextDouble() {\n return Double.parseDouble(next());\n}\n \n/**\n * Output.\n */\nPrintWriter printOut = new PrintWriter(System.out);\n \nvoid so(Object obj, boolean newLine) {\n if (newLine) printOut.println(obj);\n else printOut.print(obj);\n if (ALWAYS_FLUSH) printOut.flush();\n}\n \nvoid so(Object obj) {\n so(obj, false);\n}\n \nvoid sl(Object obj) {\n so(obj, true);\n}\n \nvoid sl() {\n sl(\"\");\n}\n \n/**\n * Main.\n */\nvoid mainGo() {\n try {\n inStream = new FileInputStream(\"src/main.in\");\n ALWAYS_FLUSH = true;\n } catch (Exception e) {\n inStream = System.in;\n }\n while (hasNext()) {\n if (MULTI_CASE) {\n int T = nextInt();\n if (T == 0) break;\n for (int i = 0; i < T; ++i) {\n go();\n }\n } else {\n go();\n }\n }\n printOut.flush();\n}\n \npublic static void main(String[] args) throws Exception {\n new Main().mainGo();\n}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 279235488,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\nimport java.io.BufferedOutputStream;\nimport java.io.BufferedReader;\nimport java.io.BufferedWriter;\nimport java.io.File;\nimport java.io.FileNotFoundException;\nimport java.io.FileReader;\nimport java.io.IOException;\nimport java.io.InputStream;\nimport java.io.InputStreamReader;\nimport java.io.OutputStream;\nimport java.io.OutputStreamWriter;\t\nimport java.io.PrintWriter;\nimport java.lang.reflect.Array;\nimport java.math.BigInteger; \nimport java.util.ArrayList;\nimport java.util.Arrays;\nimport java.util.Collection;\nimport java.util.Collections;\nimport java.util.Comparator;\nimport java.util.HashMap;\nimport java.util.HashSet;\nimport java.util.Iterator;\nimport java.util.List;\nimport java.util.Map;\nimport java.util.PriorityQueue;\nimport java.util.Queue;\nimport java.util.Scanner;\nimport java.util.Set;\nimport java.util.Stack;\nimport java.util.StringTokenizer;\nimport java.util.TreeSet;\n \n \n \n \n \npublic class Main {\n\tstatic kattio sc = new kattio();\n\tstatic PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));\n \n\tpublic static void main(String[] args) {\n\t\t//stack=new Stack<Integer>();\n\t\t//set=new HashSet<List<Integer>>();\n\t\t//f(0, new int[3]);System.out.println(set.size());\n\t\t//p1=false;\n\t\t\n\t\t\n\t\t int t=sc.nextint(); while (t-->0) {\n\t\t\t\n\t\t \n\t\t sovle(); } \n\t\t \n\t\t out.close();\n\t\t \n \n\t\t\n \n\t}\n\tstatic long[] ram,v;\n\tstatic int []size,ss;\n\tstatic int gcd(int a,int b) {\n\t\treturn b==0?a:gcd(b, a%b);\n\t}\n\tstatic void sovle2() {\n\t\tint n=sc.nextint();\n\t\tlong w=sc.nextlong();\n\t\tmap=new HashMap<Integer, List<Integer>>();\n\t\tfor (int i = 0; i < n; i++) {\n\t\t\tmap.put(i+1, new ArrayList<Integer>());\n\t\t}\n\t\tint []p=new int[n+1];\n\t\tbj=new int[n+1];\n\t\tfor (int i = 2; i <= n; i++) {\n\t\t\tp[i]=sc.nextint();\n\t\t\tmap.get(p[i]).add(i);\n\t\t}\n\t\tcnt=new int[n+1];\n\t\tf1=new int[n+1];\n\t\tf2=new int[n+1];\n\t\t\n\t\tid=0;\n\t\tdfs(1, 0);\n\t\t\n\t\tlong ge=n;long sum=0;\n\t\tlong []dis=new long [n+1];\n\t\tint []time=new int[n+1];\n\t\tfor (int i = 1; i <n; i++) {\n\t\t\tint x=sc.nextint();\n\t\t\tlong y=sc.nextlong();\n\t\t\tcnt[f1[x]]--;\n\t\t\tdis[f1[x]]+=y;\n\t\t\tif (cnt[f1[x]]==0) {\n\t\t\t\tge--;\n\t\t\t}\n\t\t\tcnt[f2[x]]--;\n\t\t\tdis[f2[x]]+=y;\n\t\t\tif (cnt[f2[x]]==0) {\n\t\t\t\tge--;\n\t\t\t}\n\t\t\tsum+=y;\n\t\t\tSystem.out.println(ge+\" \"+sum);\n\t\t\tout.print((ge*w+(2-ge)*sum)+\" \");\n\t\t}\n\t\tout.println();\n\t}\n\t\n\tstatic int []cnt,f1,f2;\n\tstatic int id;\n\tstatic void dfs(int x,int fu) {\n\t\tid++;\n\t\t\n\t\tfor (int i : map.get(x)) {\n\t\t\tcnt[id]++;\n\t\t\tf1[i]=id;\n\t\t\tdfs(i, x);\n\t\t\tcnt[id]++;\n\t\t\tf2[i]=id;\n\t\t}\n\t\t\n\t}\n\tstatic int[]bj;\n\tstatic boolean p1;\n\tstatic int[][] va,l,r;\n\tstatic int n,m;\n\tstatic void sovle() {\n\t\tn=sc.nextint();\n\t\tint q=sc.nextint();\n\t\tint []shu=new int[n+1];\n\t\tint []qian=new int[n+1];\n\t\tfor (int i = 1; i < shu.length; i++) {\n\t\t\tshu[sc.nextint()]++;\n\t\t}\n\t\tfor (int i = 1; i < shu.length; i++) {\n\t\t\tqian[i]=qian[i-1]+shu[i];\n\t\t}\n\t\tint []ans=new int[n+1];\n\t\tc=new long[n+1];\n\t\tint []va=new int[n+1];\n\t\tlist=new ArrayList<Integer>();\n\t\tint xian=n/2+1;\n\t\tint shang=n;\n\t\tfor (int i = n; i >=1; i--) {\n\t\t\tif (shang-i>i) {\n\t\t\t\tfor (int j = i+i+1; j <=shang; j++) {\n\t\t\t\t\tadd(j-i, va[j]);\n\t\t\t\t\tva[j-i]+=va[j];\n\t\t\t\t}\n\t\t\t\tshang=i+i;\n\t\t\t\t\n\t\t\t}\n\t\t\tint j = get(i, shu);\n\t\t\t//System.out.println(j);\n\t\t\tva[i]=j;\n\t\t\tadd(i, j);\n\t\t\tint l=0;int r=i;int mid;\n\t\t\twhile (l<r) {\n\t\t\t\tmid=(l+r)/2;\n\t\t\t\t\n\t\t\t\tif (qian[mid]+sum(Math.min(n, i+mid))>=xian) {\n\t\t\t\t\tr=mid;\n\t\t\t\t}else {\n\t\t\t\t\tl=mid+1;\n\t\t\t\t}\n\t\t\t}\n\t\t\tans[i]=r;\n\t\t}\n\t\t//out.println(Arrays.toString(ans));\n\t\tfor (int i = 0; i < q; i++) {\n\t\t\tout.print(ans[sc.nextint()]+\" \");\n\t\t}\n\t\tout.println();\n\t}\n\tstatic List<Integer>list;\n\t\n\tstatic int get(int x,int []shu) {\n\t\tint ans=(int) (shu[x]+sum(Math.min(n, x*2))-sum(Math.min(n, x*2-1)));\n\t\t\n\t\treturn ans;\n\t}\n \n\tstatic long pow(long x,long a) {\n\t\tif (a==0) {\n\t\t\treturn 1;\n\t\t}else {\n\t\t\tlong c=pow(x, a/2);\n\t\t\tif (a%2==0) {\n\t\t\t\treturn c*c%mod;\n\t\t\t}else {\n\t\t\t\treturn c*c%mod*x%mod;\n\t\t\t}\n\t\t}\n\t}\n\tstatic long mod = (long) (1e9+7);\n \n\tstatic void sovle1() {\n\t\t\n\t}\n \n\tstatic long []c;\n\tstatic int lowbit(int x) {\n\t\treturn x&-x;\n\t}\n\tstatic void add(int x,long v) {\n\t\twhile (x<c.length) {\n\t\t\t//System.out.println(x);\n\t\t\tc[x]+=v;\n\t\t\tx+=lowbit(x);\n\t\t}\n\t}\n\tstatic long sum(int x) {\n\t\tlong ans=0;\n\t\t//System.out.println(x+\"dasdsafadf\");\n\t\twhile (x>0) {\n\t\t\t//System.out.println(x);\n\t\t\tans+=c[x];\n\t\t\tx-=lowbit(x);\n\t\t}\n\t\treturn ans;\n\t}\n\tstatic Map<Integer, List<Integer>>map;\n\tstatic void mapinit(int chu,int n) {\n\t\tmap=new HashMap<Integer, List<Integer>>();\n\t\tfor (int i = chu; i < n+chu; i++) {\n\t\t\tmap.put(i, new ArrayList<Integer>());\n\t\t}\n\t}\n}\n \nclass kattio extends PrintWriter {\n \n\tBufferedReader r;\n\tStringTokenizer st;\n \n\tpublic kattio() {\n\t\tthis(System.in, System.out);\n\t}\n \n\tpublic kattio(InputStream i, OutputStream o) {\n\t\tsuper(o);\n\t\tr = new BufferedReader(new InputStreamReader(i));\n\t}\n \n\tpublic kattio(String in, String out) throws IOException {\n\t\tsuper(out);\n\t\tr = new BufferedReader(new FileReader(in));\n\t}\n \n\tpublic String next() {\n\t\ttry {\n\t\t\twhile (st == null || !st.hasMoreTokens()) {\n\t\t\t\tst = new StringTokenizer(r.readLine());\n\t\t\t}\n\t\t\treturn st.nextToken();\n\t\t} catch (Exception e) {\n\t\t\t// TODO: handle exception\n\t\t\treturn null;\n\t\t}\n\t}\n \n\tpublic int nextint() {\n\t\treturn Integer.parseInt(next());\n\t}\n \n\tpublic long nextlong() {\n\t\treturn Long.parseLong(next());\n\t}\n \n\tpublic double nextdouble() {\n\t\treturn Double.parseDouble(next());\n\t}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 279173352,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\n//In the name of God\n#include<bits/stdc++.h>\n/*MHN*/\nusing namespace std;\ntypedef long long int lli;\ntypedef long double ld;\ntypedef pair<lli,lli> pii;\ntypedef vector<lli> ve;\ntypedef vector<pii> vp;\nconst lli N=4e5+100;\nconst lli mod=1e9+7;//998244353;//1e9+9\nconst lli INF=1e18;\nlli power(lli x,lli y){lli res = 1;x = x % mod;while(y>0){if( y & 1 ){res = (res * x) % mod;}y = y >> 1;x = (x * x)%mod;}return res;}\nlli modinv(lli x){return power(x,mod-2);}\nlli divide(lli x,lli y){return ((x%mod) * modinv(y))%mod;}\n#define all(v) v.begin(),v.end()\n#define MIN(v) *min_element(all(v))\n#define MAX(v) *max_element(all(v))\n#define migmig ios_base::sync_with_stdio(false),cout.tie(0), cin.tie(0);\n#define read freopen(\"input.txt\", \"r\", stdin);freopen(\"output.txt\", \"w\", stdout);\n#define minheap priority_queue<pair<lli,pii>, std::vector<pair<lli,pii>>, std::greater<pair<lli,pii>>>\n#define set_preci(x) cout << fixed << setprecision(x);\n#define debug(x) cerr << (#x) << \" \" << (x) << endl\n#define MP make_pair\n#define PB push_back\n#define fi first\n#define se second\n#define SUM(v) accumulate(v.begin(),v.end(), 0LL)\n#define FOR(x,n) for(lli x = 0; x < n; x++)\n#define FORS(x,n) for(lli x = 1; x <= n; x++)\n#define TEST int TQPWIEJR; cin>>TQPWIEJR;while(TQPWIEJR--)\n#define BN(l,a) while(l){a.PB(l%2);l/=2;}\n#define lb lower_bound\n#define ub upper_bound\n#define endl \"\\n\"\n#define sep \" \"\nlli tmp;\nve t;\nint main(){\n migmig;\n TEST{\n lli n,q;\n cin>>n>>q;\n ve v(n);\n FOR(i,n){\n cin>>v[i];\n }\n while(q--){\n lli x;\n cin>>x;\n t.clear();\n FOR(i,n){\n t.PB(v[i]%x);\n }\n sort(all(t));\n cout<<t[n/2]<<sep;\n }\n cout<<endl;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 280749662,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n#define int long long\n#define endl \"\\n\"\t\t\n#define run freopen(\"input.txt\",\"r\",stdin);freopen(\"output.txt\",\"w\",stdout);\n#define bismillah ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);\n#define pii pair<int,int>\n#define f first\n#define s second\n \nint MOD=1e9+7;\nint db[]={0,0,1,-1};\nint dk[]={1,-1,0,0};\n \nint pwr(int a,int b){\n if(b==0)return 1;\n int tmp=pwr(a,b/2);\n tmp=(tmp*tmp)%MOD;\n if(b%2)tmp=(tmp*a)%MOD;\n return tmp%MOD;\n}\n \nvector<int> vt(1e5+5);\nint x;\n \nbool cmp(int a,int b){\n return a%x<=b%x;\n}\nsigned main() \n{\n // run \n bismillah\n \n int t;cin>>t;\n while(t--){\n int n,q;cin>>n>>q;\n for(int i=0;i<n;i++)\n cin>>vt[i];\n vector<int>dp(n+1,-1);\n while(q--){\n cin>>x;\n if(dp[x]!=-1){\n cout<<dp[x]<<\" \";\n continue;\n }\n sort(vt.begin(),vt.begin()+n,cmp);\n int ans;\n if(n&1)ans=vt[((n+1)/2)-1]%x;\n else ans=vt[((n+2)/2)-1]%x;\n dp[x]=ans;\n cout<<ans<<\" \";\n }\n cout<<endl;\n }\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 291286346,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\nfrom collections import Counter\nfor _ in range(int(input())):\n n, Q = map(int, input().split())\n medc = (n+1)//2 if n%2 else (n//2 + 1)\n a = Counter(map(int, input().split()))\n dp = [-1]*n\n dp2 = [-1]*n\n \n for q in range(Q):\n x = int(input())\n if not dp2[x-1]+1:\n if not dp[x-1]+1:\n l = [0]*x\n for i in a: l[i%x]+=a[i]\n for i in range(1, x): l[i] += l[i-1]\n dp[x-1] = l\n l = 0\n h = x-1\n while l < h:\n mid = (l+h)//2\n c = dp[x-1][mid]\n if c < medc: l = mid+1\n else: h = mid\n dp2[x-1] = h\n print(dp2[x-1], end = \" \")\n print()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 282827411,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\ndef findMid(l, h, a, x, medc):\n while l < h:\n mid = (l+h)//2\n c = sum(i%x <= mid for i in a)\n if c < medc: l = mid+1\n else: h = mid\n return h\n \nfor _ in range(int(input())):\n n, Q = map(int, input().split())\n medc = (n+1)//2 if n%2 else (n//2 + 1)\n a = list(map(int, input().split()))\n for q in range(Q):\n x = int(input())\n print(findMid(0, x-1, a, x, medc), end = \" \")\n print()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 282822294,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\nimport java.util.Scanner;\nimport java.util.Arrays;\n \npublic class Solution\n{\n int check(int l, int arr[], int x)\n {\n for(int i=0;i<l;i++)\n {\n arr[i]=arr[i]%x;\n }\n Arrays.sort(arr);\n return arr[l/2];\n }\n public static void main(String args[])\n {\n Scanner k=new Scanner(System.in);\n Solution ob=new Solution();\n int n=k.nextInt();\n for(int i=0;i<n;i++)\n {\n int l=k.nextInt();\n int q=k.nextInt();\n int arr1[]=new int[l];\n for(int j=0;j<l;j++)\n {\n arr1[j]=k.nextInt();\n }\n for(int m=0;m<q;m++)\n {\n int arr[]=new int[l];\n for(int c=0;c<l;c++)\n {\n arr[c]=arr1[c];\n }\n if(l==1)\n {\n System.out.println(arr[0]%k.nextInt());\n continue;\n }\n else if(l==2)\n {\n int x=k.nextInt();\n arr[0]=arr[0]%x;\n arr[1]=arr[1]%x;\n System.out.println(((arr[0])>(arr[1]))?arr[0]:arr[1]);\n continue;\n }\n else\n System.out.println(ob.check(l,arr,k.nextInt()));\n }\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 279211174,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\nimport java.util.Scanner;\nimport java.util.Arrays;\n \npublic class Solution\n{\n int check(int l, int arr[], int x)\n {\n for(int i=0;i<l;i++)\n {\n arr[i]=arr[i]%x;\n }\n Arrays.sort(arr);\n return arr[l/2];\n }\n public static void main(String args[])\n {\n Scanner k=new Scanner(System.in);\n Solution ob=new Solution();\n int n=k.nextInt();\n for(int i=0;i<n;i++)\n {\n int l=k.nextInt();\n int q=k.nextInt();\n int arr1[]=new int[l];\n for(int j=0;j<l;j++)\n {\n arr1[j]=k.nextInt();\n }\n for(int m=0;m<q;m++)\n {\n int arr[]=new int[l];\n for(int c=0;c<l;c++)\n {\n arr[c]=arr1[c];\n }\n if(l==1)\n {\n System.out.println(arr[0]%k.nextInt());\n continue;\n }\n if(l==2)\n {\n int x=k.nextInt();\n arr[0]=arr[0]%x;\n arr[1]=arr[1]%x;\n System.out.println(((arr[0])>(arr[1]))?arr[0]:arr[1]);\n continue;\n }\n System.out.println(ob.check(l,arr,k.nextInt()));\n }\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 279204192,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\nfor _ in range(int(input())):\n n, m = map(int, input().split())\n a = list(map(int, input().split()))\n c = [0] * (n + 1)\n \n for i in range(n):\n c[a[i]] += 1\n \n for i in range(1, n + 1):\n c[i] += c[i - 1]\n \n res = [0] * (n + 1)\n \n for x in range(1, n + 1):\n l, r = 0, x\n while l < r:\n mid = (l + r) // 2\n cnt = c[mid]\n for k in range(1, n // x + 1):\n cnt += c[min(k * x + mid, n)] - c[k * x - 1]\n if cnt - 1 >= n // 2:\n r = mid\n else:\n l = mid + 1\n res[x] = l\n \n for _ in range(m):\n x = int(input())\n print(res[x])\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 287173535,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\nimport sys\n \ndef calculate_median(freq, n):\n count = 0\n prev_count = 0\n prev_index = -1\n for i in range(len(freq)):\n prev_count = count\n prev_index = i\n count += freq[i]\n if n % 2 == 1:\n if count >= (n + 1) // 2:\n return i\n else:\n if count >= (n + 2) // 2:\n if prev_count < (n + 2) // 2 - 1:\n return i\n else:\n return min(i, prev_index)\n \ndef solve():\n n, q = map(int, sys.stdin.readline().split())\n arr = sorted(list(map(int, sys.stdin.readline().split())))\n \n for _ in range(q):\n x = int(sys.stdin.readline())\n freq = [0] * x\n for num in arr:\n freq[num % x] += 1\n median = calculate_median(freq, n)\n print(median, end=\" \")\n print()\n \ndef main():\n t = int(sys.stdin.readline())\n for _ in range(t):\n solve()\n \nif __name__ == \"__main__\":\n main()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 279162987,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\n#pragma GCC optimize(2)\n#pragma GCC optimize(3,\"Ofast\",\"inline\")\n#include<bits/stdc++.h>\nusing namespace std;\n#ifdef LOCAL\n#include \"D:\\Users\\TauLee\\OneDrive\\Mine\\c++\\debug.h\"\n#else\n#define debug(...) 42;\n#define endl '\\n'\n#endif\n \nvoid solve(int _) {\n int n, q, x;\n map<int, int>ans;\n cin >> n >> q;\n vector<int>a(n);\n for (auto &c : a)cin >> c;\n sort(a.begin(), a.end());\n \n debug(a)\n auto check = [&](int mid) {\n int cnt = 0;\n for (int i = 0; i < INT_MAX; ++i) {\n if (i * x > a[n - 1])break;\n int l = upper_bound(a.begin(), a.end(), i * x - 1) - a.begin();\n int r = upper_bound(a.begin(), a.end(), i * x + mid) - a.begin();\n cnt += r - l;\n }\n return cnt >= (n + 2) / 2;\n };\n \n while (q--) {\n cin >> x;\n if (ans.count(x)) {\n cout << ans[x] << endl;\n continue;\n }\n int l = 0, r = INT_MAX;\n while (l < r) {\n int mid = (r + l) / 2;\n if (check(mid)) {\n r = mid;\n }\n else {\n l = mid + 1;\n }\n }\n ans[x] = l;\n cout << ans[x] << \" \";\n }\n}\nsigned main()\n{\n ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);\n int _ = 1;\n cin >> _;\n while (_--) {\n solve(_);\n cout << endl;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 279915012,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \n#define lep(i, l, r) for(int i = (l); i <= (r); i ++)\n#define rep(i, l, r) for(int i = (l); i >= (r); i --)\n#define Lep(i, l, r) for(int i = (l); i < (r); i ++)\n#define debug(...) fprintf (stderr, __VA_ARGS__)\n \nusing i64 = long long;\nusing ld = long double;\nusing LL = long long;\nusing ui64 = unsigned long long;\n \nconst int P = 998244353;\ninline int mod(int x) { return x + (x >> 31 & P); }\ninline void pls(int &x, int y) { x = mod(x + y - P); }\ninline void sub(int &x, int y) { x = mod(x - y); }\ninline int add(int x, int y) { return mod (x + y - P); }\ninline int dec(int x, int y) { return mod (x - y); }\ntemplate <typename T> inline void ckmin(T &x, T y) { if (x > y) x = y; }\ntemplate <typename T> inline void ckmax(T &x, T y) { if (x < y) x = y; }\n \ninline int power(int x, int k) {\n\tint res = 1;\n\twhile(k) {\n\t\tif (k & 1) res = 1ll * res * x % P;\n\t\tx = 1ll * x * x % P; k >>= 1;\n\t} return res;\n}\n \nvoid solve() {\n\tint n, q;\n\tcin >> n >> q;\n\t//bt.init (n);\n\tvector<int> a(n), b(n), c(n + 1);\n\tfor (int &x : a) cin >> x, c[x] ++;\n\tfor (int i = 1; i <= n; i ++) c[i] += c[i - 1];\n \n\tint B = pow(n, 0.6666666);\n\tvector<int> ans1(B + 1);\n \n\tfor (int x = 1; x <= B; x ++) {\n\t\tfor (int i = 0; i < n; i ++) {\n\t\t\tb[i] = a[i] % x;\n\t\t}\n\t\t//sort (b.begin(), b.end());\n\t\t//\n\t\tnth_element(b.begin(), b.begin() + (n >> 1), b.end());\n\t\tans1[x] = b[n / 2];\n\t}\n\t//cerr << B << endl;\n \n\tunordered_map<int, int> mp;\n \n\twhile (q --) {\n\t\tint x;\n\t\tcin >> x;\n\t\tif (x <= B) cout << ans1[x] << ' ';\n\t\telse {\n\t\t\tif (mp.count(x) != 0) {\n\t\t\t\tcout << mp[x] << ' ';\n\t\t\t\tcontinue;\n\t\t\t}\n\t\t\tint ans = 0;\n\t\t\tint l = 0, r = x - 1, res = n;\n \n\t\t\tauto check = [&] (int mid) -> int {\n\t\t\t\tint sum = 0;\n\t\t\t\tfor (int k = 0; k * x <= n; k ++) {\n\t\t\t\t\tint l = k * x;\n\t\t\t\t\tint r = k * x + mid;\n\t\t\t\t\tif (r > n) r = n;\n\t\t\t\t\t//cerr << k << endl;\n\t\t\t\t\tsum += c[r];\n\t\t\t\t\tif (l) sum -= c[l - 1];\n\t\t\t\t}\n\t\t\t\treturn sum >= n / 2 + 1;\n\t\t\t} ;\n \n\t\t\twhile (l <= r) {\n\t\t\t\tint mid = (l + r) >> 1;\n\t\t\t\t//cerr << mid << endl;\n\t\t\t\tif (check (mid)) r = mid - 1, res = mid;\n\t\t\t\telse l = mid + 1;\n\t\t\t}\n \n\t\t\tmp[x] = res;\n\t\t\tcout << res << ' ';\n\t\t}\n\t}\n\tcout << '\\n';\n}\n \nint main() {\n\tios :: sync_with_stdio(false);\n\tcin.tie(0); cout.tie(0);\n\tint Case;\n\tcin >> Case;\n\twhile (Case --) solve();\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 279141023,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.math.*;\nimport java.util.*;\n \nimport static java.lang.Math.min;\n \npublic class H_2008 {\n \n private static final int MAX = 200005;\n \n public static void main(String[] args) throws IOException {\n final var cin = new Reader();\n final var sb = new StringBuilder();\n int t = cin.nextInt();\n while (t-- > 0) {\n int n = cin.nextInt();\n int q = cin.nextInt();\n int[] arr = new int[n + 1];\n for (int i = 0; i < n; i++) {\n int x = cin.nextInt();\n arr[x]++;\n }\n for (int i = 1; i <= n; i++) arr[i] += arr[i - 1];\n int medium = n/2 + 1;\n while (q-- > 0) {\n int x = cin.nextInt();\n int l = 0, r = x - 1;\n while (l < r) {\n int m = (l + r) / 2;\n int items = 0;\n for (int i = 0; i <= n; ) {\n int upperBound = arr[min(i + m, n)];\n if (i == 0) items += upperBound;\n else items += upperBound - arr[i - 1];\n i += x;\n }\n if(items >= medium) r = m;\n else l= m+1;\n }\n sb.append(l).append(\" \");\n }\n sb.append(\"\\n\");\n }\n \n System.out.println(sb);\n }\n \n static class Reader {\n final private int BUFFER_SIZE = 1 << 16;\n private DataInputStream din;\n private byte[] buffer;\n private int bufferPointer, bytesRead;\n \n public Reader() {\n din = new DataInputStream(System.in);\n buffer = new byte[BUFFER_SIZE];\n bufferPointer = bytesRead = 0;\n }\n \n public Reader(String file_name) throws IOException {\n din = new DataInputStream(\n new FileInputStream(file_name));\n buffer = new byte[BUFFER_SIZE];\n bufferPointer = bytesRead = 0;\n }\n \n public String readLine() throws IOException {\n int x = 256;\n byte[] buf = new byte[x];\n int cnt = 0, c;\n while ((c = read()) != -1) {\n if (c == '\\n') {\n if (cnt != 0) {\n break;\n } else {\n continue;\n }\n }\n if (cnt >= x) {\n x = x * 2;\n byte[] temp = new byte[x];\n System.arraycopy(buf, 0, temp, 0, cnt);\n buf = temp;\n }\n buf[cnt++] = (byte) c;\n }\n return new String(buf, 0, cnt);\n }\n \n public int nextInt() throws IOException {\n int ret = 0;\n byte c = read();\n while (c <= ' ') {\n c = read();\n }\n boolean neg = (c == '-');\n if (neg)\n c = read();\n do {\n ret = ret * 10 + c - '0';\n } while ((c = read()) >= '0' && c <= '9');\n \n if (neg)\n return -ret;\n return ret;\n }\n \n public long nextLong() throws IOException {\n long ret = 0;\n byte c = read();\n while (c <= ' ')\n c = read();\n boolean neg = (c == '-');\n if (neg)\n c = read();\n do {\n ret = ret * 10 + c - '0';\n } while ((c = read()) >= '0' && c <= '9');\n if (neg)\n return -ret;\n return ret;\n }\n \n public double nextDouble() throws IOException {\n double ret = 0, div = 1;\n byte c = read();\n while (c <= ' ')\n c = read();\n boolean neg = (c == '-');\n if (neg)\n c = read();\n \n do {\n ret = ret * 10 + c - '0';\n } while ((c = read()) >= '0' && c <= '9');\n \n if (c == '.') {\n while ((c = read()) >= '0' && c <= '9') {\n ret += (c - '0') / (div *= 10);\n }\n }\n \n if (neg)\n return -ret;\n return ret;\n }\n \n private void fillBuffer() throws IOException {\n bytesRead = din.read(buffer, bufferPointer = 0,\n BUFFER_SIZE);\n if (bytesRead == -1)\n buffer[0] = -1;\n }\n \n private byte read() throws IOException {\n if (bufferPointer == bytesRead)\n fillBuffer();\n return buffer[bufferPointer++];\n }\n \n public void close() throws IOException {\n if (din == null)\n return;\n din.close();\n }\n }\n \n static final class Pair implements Comparable<Pair> {\n int x;\n int y;\n \n Pair(int x, int y) {\n this.x = x;\n this.y = y;\n }\n \n @Override\n public String toString() {\n return \"(\" + x + \", \" + y + \")\";\n }\n \n @Override\n public int hashCode() {\n return x + y + (y << 16);\n }\n \n @Override\n public boolean equals(Object obj) {\n if (this == obj)\n return true;\n if (obj == null)\n return false;\n if (getClass() != obj.getClass())\n return false;\n Pair other = (Pair) obj;\n return x == other.x && y == other.y;\n }\n \n @Override\n public int compareTo(Pair o) {\n if (this.x < o.x) return -1;\n if (this.x > o.x) return 1;\n if (this.y < o.y) return -1;\n if (this.y > o.y) return 1;\n return 0;\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 285003347,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Sakurako's Test\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Sakurako will soon take a test. The test can be described as an array of integers $n$ and a task on it:\n\nGiven an integer $x$, Sakurako can perform the following operation any number of times:\n\n * Choose an integer $i$ ($1\\le i\\le n$) such that $a_i\\ge x$; * Change the value of $a_i$ to $a_i-x$. \n\nUsing this operation any number of times, she must find the minimum possible median$^{\\text{โ}}$ of the array $a$.\n\nSakurako knows the array but does not know the integer $x$. Someone let it slip that one of the $q$ values of $x$ will be in the next test, so Sakurako is asking you what the answer is for each such $x$.\n\n$^{\\text{โ}}$The median of an array of length $n$ is the element that stands in the middle of the sorted array (at the $\\frac{n+2}{2}$-th position for even $n$, and at the $\\frac{n+1}{2}$-th for odd)\n\nInput: The first line contains one integer $t$ ($1\\le t\\le 10^4$) โ the number of test cases.\n\nThe first line of each test case contains two integers $n$ and $q$ ($1\\le n,q\\le 10^5$) โ the number of elements in the array and the number of queries.\n\nThe second line of each test case contains $n$ integers $a_1, a_2, \\dots, a_n$ ($1\\le a_i\\le n$) โ the elements of the array.\n\nThe following $q$ lines each contain one integer $x$ ($1\\le x\\le n$).\n\nIt is guaranteed that the sum of $n$ across all test cases does not exceed $10^5$. The same guarantee applies to the sum of $q$ across all test cases.\n\nOutput: For each test case, output $q$ integers โ the answer for each query.\n\nExamples:\ninput:\n2\n5 5\n1 2 3 4 5\n1\n2\n3\n4\n5\n6 3\n1 2 6 4 1 3\n2\n1\n5\noutput:\n0 1 1 1 2 \n1 0 2\n\nNote: \n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Main {\n static final int N = 100010;\n static long[] a = new long[N], cnt = new long[N];\n \n public static void main(String[] args) throws IOException {\n FastScanner f = new FastScanner();\n PrintWriter w = new PrintWriter(System.out);\n int T = f.nextInt();\n while (T-- > 0) {\n int n = f.nextInt();\n int q = f.nextInt();\n cnt = new long[n + 1];\n a = new long[n + 1];\n for (int i = 1; i <= n; i++) {\n a[i] = f.nextInt();\n cnt[(int) a[i]]++;\n }\n for (int i = 1; i <= n; i++) cnt[i] += cnt[i - 1];\n while (q-- > 0) {\n int x = f.nextInt();\n if (x == 1) w.print(0 + \" \");\n else {\n int l = 0, r = x - 1;\n while (l < r) {\n int mid = l + r >> 1;\n int sum = 0;\n int l1 = 0, r1 = mid;\n while (l1 <= n) {\n int l2 = l1 - 1;\n int r2 = Math.min(r1, n);\n sum += cnt[r2] - (l2 < 0 ? 0 : cnt[l2]);\n l1 += x;\n r1 += x;\n }\n if (sum >= (n + 2) / 2) r = mid;\n else l = mid + 1;\n }\n w.print(l + \" \");\n }\n }\n w.println();\n }\n w.flush();\n }\n \n private static class FastScanner {\n final private int BUFFER_SIZE = 1 << 16;\n private DataInputStream din;\n private byte[] buffer;\n private int bufferPointer, bytesRead;\n \n private FastScanner() throws IOException {\n din = new DataInputStream(System.in);\n buffer = new byte[BUFFER_SIZE];\n bufferPointer = bytesRead = 0;\n }\n \n private short nextShort() throws IOException {\n short ret = 0;\n byte c = read();\n while (c <= ' ') c = read();\n boolean neg = (c == '-');\n if (neg) c = read();\n do ret = (short) (ret * 10 + c - '0');\n while ((c = read()) >= '0' && c <= '9');\n if (neg) return (short) -ret;\n return ret;\n }\n \n private int nextInt() throws IOException {\n int ret = 0;\n byte c = read();\n while (c <= ' ') c = read();\n boolean neg = (c == '-');\n if (neg) c = read();\n do ret = ret * 10 + c - '0';\n while ((c = read()) >= '0' && c <= '9');\n if (neg) return -ret;\n return ret;\n }\n \n public long nextLong() throws IOException {\n long ret = 0;\n byte c = read();\n while (c <= ' ') c = read();\n boolean neg = (c == '-');\n if (neg) c = read();\n do ret = ret * 10 + c - '0';\n while ((c = read()) >= '0' && c <= '9');\n if (neg) return -ret;\n return ret;\n }\n \n private char nextChar() throws IOException {\n byte c = read();\n while (c <= ' ') c = read();\n return (char) c;\n }\n \n private String nextString() throws IOException {\n StringBuilder ret = new StringBuilder();\n byte c = read();\n while (c <= ' ') c = read();\n do {\n ret.append((char) c);\n } while ((c = read()) > ' ');\n return ret.toString();\n }\n \n private void fillBuffer() throws IOException {\n bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);\n if (bytesRead == -1) buffer[0] = -1;\n }\n \n private byte read() throws IOException {\n if (bufferPointer == bytesRead) fillBuffer();\n return buffer[bufferPointer++];\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2008H",
"submission_id": 287452966,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n#define ll long long\n#define all(c) c.begin(),c.end()\n#define sz(c) int(c.size())\n#define tr(a,it) for(auto it=a.begin(); it!=a.end(); it++)\n#define present(c, x)(c.find(x) != c.end())\n#define cpresent(c, x)(find(all(c), x) != c.end())\n#define vi vector<int>\n#define vll vector<ll>\n#define vvll vector<vll>\n#define vpii vector<pair<int,int>>\n#define vpll vector<pair<ll,ll>>\n#define inf 10e9\n#define MOD 1000000007\n#define loop(i,s,e) for(ll i=s; i<e; i++)\n#define revloop(i,s,e) for(ll i=s; i>=e; i--)\n \nusing namespace std;\n \nvoid print(vll a){for(auto it: a) cout << it<<\" \"; cout << '\\n';}\nvoid print(set<ll> a){for(auto it: a) cout << it<<\" \"; cout << '\\n';}\n \nvoid solve() {\n int n,m; cin >> n >> m;\n string s; cin >> s;\n vi ind(m); loop(i,0,m) cin >> ind[i];\n string c; cin >> c;\n sort(all(ind)); sort(all(c));\n int j = n-1, k=0;\n loop(i,0,m){\n if(i<m && ind[i+1]==ind[i] ){\n if(j>=0 && j<m) s[ind[i]-1] = c[j];\n j--;\n }\n else{\n if(k>=0 && k<m) s[ind[i]-1] = c[k];\n k++;\n }\n }\n cout << s << endl;\n}\n \nint main() {\n ios::sync_with_stdio(0); cin.tie(0);\n int t;\n cin >> t;\n while (t--) {\n solve();\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 272074488,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n// ifstream fin(\"C:\\\\Users\\\\Bia\\\\Desktop\\\\vscode\\\\kattis\\\\i.txt\");\n// ofstream fout(\"C:\\\\Users\\\\Bia\\\\Desktop\\\\vscode\\\\kattis\\\\o.txt\");\n \ntypedef long long ll;\n \nint main()\n{\n int ca;\n cin >> ca;\n \n for (int ih = 0; ih < ca; ih += 1)\n {\n int n, m;\n cin >> n >> m;\n string s, c;\n vector<int> indx(m);\n deque<char> let(m);\n vector<char> res;\n \n cin >> s;\n for (int i = 0; i < m; i += 1)\n {\n cin >> indx[i];\n }\n \n cin >> c;\n for (int i = 0; i < m; i += 1)\n {\n let[i] = c[i];\n }\n \n sort(indx.begin(), indx.end());\n sort(let.begin(), let.end());\n \n for (int i = 1; i < m; i += 1)\n {\n if (indx[i] == indx[i - 1])\n {\n let.pop_back();\n indx[i] = -1;\n }\n }\n for (int i = 0; i < m; i += 1)\n {\n if (indx[i] != -1)\n {\n s[indx[i] - 1] = let[0];\n let.pop_front();\n }\n }\n \n cout << s << endl;\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267074267,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\n#include<iostream>\nusing namespace std;\n#include<vector>\n#include<math.h>\n#include<algorithm>\n#include<queue>\n#include<unordered_map>\n#include<string>\n#include<limits.h>\n#include <utility>\n \nstring solve(string& a, string& b, vector<int>& vec) {\n // Priority queue to get the smallest character first\n priority_queue<char, vector<char>, greater<char>> ch;\n for (char c : b) {\n ch.push(c);\n }\n \n // Vector to keep track of which positions in `a` have been visited/modified\n vector<bool> visited(a.size(), false);\n \n for (int i = 0; i < vec.size(); i++) {\n int pos = vec[i] - 1; // Adjust index since vec contains 1-based indices\n if (!visited[pos]) {\n a[pos] = ch.top();\n ch.pop();\n visited[pos] = true;\n } else {\n // Keep the minimum character if the position is already visited\n a[pos] = min(a[pos], ch.top());\n ch.pop();\n }\n }\n \n // Create a reversed version of the string `a`\n string reversed_a = a;\n reverse(reversed_a.begin(), reversed_a.end());\n \n // Return the lexicographically smaller string between `a` and `reversed_a`\n return min(a, reversed_a);\n}\n \nint main() {\n int n;\n cin >> n;\n \n for (int i = 0; i < n; i++) {\n int a_len, b_len;\n cin >> a_len >> b_len;\n \n string a;\n for (int j = 0; j < a_len; j++) {\n char ch;\n cin >> ch;\n a += ch;\n }\n \n vector<int> vec(b_len);\n for (int j = 0; j < b_len; j++) {\n cin >> vec[j];\n }\n \n string b;\n for (int j = 0; j < b_len; j++) {\n char ch;\n cin >> ch;\n b += ch;\n }\n \n cout << solve(a, b, vec) << endl;\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267070360,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\nfrom collections import Counter\nfrom collections import defaultdict\nfrom collections import deque\nimport math\nimport heapq\nt = int(input())\n \nfor i in range(t):\n n, m = input().split()\n s = input()\n inds = list(map(int, input().split()))\n c = input()\n c = sorted(c)\n inds.sort()\n #print(c)\n #print(inds)\n res = []\n left, right = 0, len(c) - 1\n let_left, let_right = 0, len(c) - 1\n while left <= right and let_left <= let_right:\n while left + 1 < len(c) and inds[left] == inds[left + 1] and right > left:\n res.append([c[let_right], inds[left]])\n let_right -= 1\n left += 1\n if res and res[-1][1] == inds[left]:\n res[-1][0] = min(c[let_left], res[-1][0])\n else:\n res.append([c[let_left], inds[left]])\n left += 1\n let_left += 1\n s = list(s)\n for i in range(len(s) - 1, -1, -1):\n if res[-1][1] - 1 == i:\n s[i] = res[-1][0]\n res.pop()\n print(''.join(s))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 269495252,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\nt = int(input())\nres=[]\nfor _ in range(t):\n n, m = map(int, input().split())\n s = input().strip()\n ind = list(map(int, input().split()))\n ind=set(ind)\n c = input()\n c=sorted(c)\n updates = []\n i=0\n for e in ind:\n updates.append((e, c[i]))\n i=i+1\n result = list(s)\n \n # Apply updates in sorted order\n for index, character in updates:\n result[index - 1] = character \n \n res.append(''.join(result).strip())\nfor i in range(len(res)):\n if i ==len(res)-1:\n print(res[i][::-1])\n else:\n print(res[i])\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267061337,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\nfor _ in range(int(input())):\n n, m = map(int,input().split())\n s = list(input())\n idx = sorted(set(map(int,input().split())))\n min_i = idx[0]\n idx = list(map(lambda x: x - min_i, idx))\n c = sorted(input())\n \n for ci, i in enumerate(idx):\n s[i] = c[ci] \n print(''.join(s))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267021771,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\nimport java.io.*;\nimport java.util.*;\n \npublic class Update_Queries{\n static FastReader sc = new FastReader();\n public static void main(String[] args){\n int t = sc.nextInt();\n while (t-- > 0){\n solve();\n }\n }\n \n public static void solve(){\n int n = sc.nextInt();\n int m = sc.nextInt();\n String s = sc.next();\n StringBuilder res = new StringBuilder();\n res.append(s);\n \n int[] indices = new int[m];\n for (int i = 0; i < m; i += 1){\n indices[i] = sc.nextInt();\n }\n String up = sc.next();\n int[] update = new int[26];\n char[] chs = up.toCharArray();\n for (int i = 0; i < m; i += 1){\n update[chs[i] - 'a'] += 1;\n }\n Arrays.sort(indices);\n \n int l = 0, r = 25;\n for (int i = 0; i < m - 1; i += 1){\n while (update[r] == 0){\n r -= 1;\n }\n while (update[l] == 0){\n l += 1;\n }\n int curr = indices[i] - 1;\n int next = indices[i + 1] - 1;\n if (curr == next){\n res.setCharAt(curr, (char)('a' + r));\n update[r] -= 1;\n } else{\n res.setCharAt(curr, (char)('a' + l));\n update[l] -= 1;\n while (update[l] == 0){\n l += 1;\n }\n }\n }\n res.setCharAt(indices[m - 1] - 1, (char)('a' + l));\n \n System.out.println(res.toString());\n }\n}\n \n \nclass FastReader {\n \n BufferedReader br;\n StringTokenizer st;\n \n public FastReader() {\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n \n public FastReader(String fileName) throws IOException {\n br = new BufferedReader(new FileReader(fileName));\n }\n \n String next() {\n while (st == null || !st.hasMoreElements()) {\n try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n String nextLine() {\n String str = \"\";\n try {\n str = br.readLine();\n } catch (IOException e) {\n e.printStackTrace();\n }\n return str;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 271043317,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\n// \n// Author : Swarnadip Kar\n// IIT Bhilai\n// \n \nimport java.lang.*; // imports everything\nimport java.util.*;\n \npublic class C_Update_Queries\n{\n //static ArrayList<Integer> a=new ArrayList<>();\n //static int max;\n public static void main (String[] args) throws java.lang.Exception\n {\n Scanner sc=new Scanner(System.in);\n int caseNo,i,j,N;\n N=sc.nextInt();\n for(caseNo=0;caseNo<N;caseNo++)\n {\n int n=sc.nextInt();\n int m=sc.nextInt();\n \n String str=sc.next();\n char a[]=new char[n];\n for(i=0;i<n;i++)\n {\n a[i]=str.charAt(i);\n }\n \n ArrayList<Integer> ind=new ArrayList<>();\n for(i=0;i<m;i++)\n {\n ind.add(sc.nextInt());\n }\n \n str=sc.next();\n char c[]=new char[m];\n for(i=0;i<m;i++)\n {\n c[i]=str.charAt(i);\n }\n \n Collections.sort(ind);\n //Collections.sort(a,Collections.reverseOrder());\n Arrays.sort(c,0,m);\n for(i=1;i<m;i++)\n {\n if(ind.get(i).equals(ind.get(i-1)))\n {\n ind.remove(i);\n m--;\n }\n }\n int k=0;\n for(i=0;i<m;i++)\n {\n a[ind.get(i)-1]=c[k++];\n }\n for(i=0;i<n;i++)\n {\n System.out.print(a[i]);\n }\n System.out.println();\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267020332,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\nusing ll = long long;\n \nvoid solve(){\n int n, m;\n string s, c, final = \"s\";\n cin >> n >> m >> s;\n int ind[m+2];\n for(int i = 0; i < m; i++){\n cin >> ind[i];\n }\n cin >> c;\n \n sort(ind, ind + m);\n sort(c.begin(), c.end());\n for(int i = 0; i < m; i++){\n if(ind[i] == ind[i+1]){\n c.erase(c.begin() + (c.length()-1));\n }\n }\n \n int k = 0;\n for(int i = 0; i < m; i++){\n if(ind[i] > n) break;\n if(ind[i] != ind[i+1]){\n s[ind[i]-1] = c[k++];\n }\n }\n \n cout << s << \"\\n\";\n}\n \nint main()\n{\n ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);\n int t; cin>>t;\n while(t--) solve();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267065007,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \n \nint main() {\n\tios_base::sync_with_stdio(false);\n\tcin.tie(NULL);\n\tint t;\n\tcin >> t;\n\twhile (t--) {\t\n\t\tint n, m, y;\n\t\tcin >> n >> m;\n\t\tint pos = m;\n\t\tvector<char> s, c;\n\t\tvector<bool> ex(n, false);\n\t\tstring x;\n\t\tvector<int> ind;\n\t\tcin >> x;\n\t\tfor (auto i : x) s.push_back(i);\n\t\tfor (int i = 0; i < m; i++) {\n\t\t\tcin >> y;\n\t\t\tif (!ex[y]) {\n\t\t\t\tind.push_back(y);\n\t\t\t\tex[y] = true;\n\t\t\t} else pos--;\n\t\t}\n\t\tcin >> x;\n\t\tfor (auto i : x) c.push_back(i);\n\t\tsort(c.begin(), c.end());\n\t\tsort(ind.begin(), ind.end());\n\t\tfor (int i = pos - 1; i >= 0; i--) {\n\t\t\ts[ind[i] - 1] = c[i];\n\t\t}\n\t\tfor (auto i : s) cout << i;\n\t\tcout << '\\n';\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267469472,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\nimport java.util.*;\n \npublic class Main {\n \n public static void solve(Scanner sc) {\n int n = sc.nextInt();\n int m = sc.nextInt();\n String s = sc.next();\n int[] a = new int[m];\n Set<Integer> st = new HashSet<>();\n \n for (int i = 0; i < m; i++) {\n a[i] = sc.nextInt();\n }\n for (int ele : a) {\n st.add(ele - 1);\n }\n String t = sc.next();\n char[] tArr = t.toCharArray();\n Arrays.sort(tArr);\n int ind = 0;\n StringBuilder sb = new StringBuilder(s);\n \n for (int ele : st) {\n sb.setCharAt(ele, tArr[ind++]);\n }\n System.out.println(sb.toString());\n }\n \n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n while (t-- > 0) {\n solve(sc);\n }\n sc.close();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267854257,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\n//package org.example.div2;\n \nimport java.util.Arrays;\nimport java.util.HashSet;\nimport java.util.Scanner;\nimport java.util.Set;\n \npublic class UpdateQueries {\n \n public static void main(String[] args) {\n \n Scanner sc = new Scanner(System.in);\n solveT(sc);\n \n \n }\n public static void solveT(Scanner sc){\n \n int t = sc.nextInt();\n \n while (t-- > 0){\n \n int n = sc.nextInt();\n int m = sc.nextInt();\n \n Set<Integer> set = new HashSet<>();\n String a = sc.next();\n \n for(int i = 0 ; i < m ; i++){\n set.add(sc.nextInt()-1);\n }\n \n String b = sc.next();\n \n char [] ch = b.toCharArray();\n Arrays.sort(ch);\n \n char [] imp = a.toCharArray();\n \n int idx = 0;\n \n for(int i : set){\n imp[i] = ch[idx++];\n }\n \n String ans = new String(imp);\n System.out.println(ans);\n \n }\n \n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 268013139,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\ndef sort_string(s):\n chars = list(s)\n n = len(chars)\n for i in range(n):\n for j in range(0, n-i-1):\n if chars[j] > chars[j+1]:\n chars[j], chars[j+1] = chars[j+1], chars[j]\n return ''.join(chars)\nt = int(input())\nfor i in range(t):\n list1=list(map(int, input().strip().split()))\n b=input()\n list2=list(map(int, input().strip().split()))\n c=input()\n d =sort_string(c)\n k=0\n for o in range(1,len(b)+1):\n if o in list2:\n print(d[k],end=\"\")\n k+=1\n else:\n print(b[o-1],end=\"\")\n print('\\n',end='')\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 272496254,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\nfor _ in range(int(input())):\n x,y=map(int, input().split())\n a=list(input())\n ind=sorted(list(map(int, input().split())))\n b=sorted(list(input()))\n ind.append(-1)\n for i in range(y):\n if ind[i]==ind[i+1]:\n b=[*b[:i],b[-1],*b[i:-1]]\n b=list(b)\n for i in range(y):\n a[ind[i]-1]=b[i]\n print(''.join(a))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267122177,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\nfor t in range(int(input())):\n n,m=map(int, input().split())\n s=list(input())\n ind=list(map(int, input().split()))\n c=list(input())\n ind.sort()\n \n find=[]\n x=sorted(c)\n for i in ind:\n if i not in find:\n find.append(i)\n \n d=len(ind)-len(find)\n fc=x[:len(x)-d]\n \n i=0\n for j in find:\n s[j-1]=fc[i]\n i+=1\n \n print(''.join(s))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267066635,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nint main() {\n int t;\n cin >> t;\n while (t--) {\n int n, m;\n cin >> n >> m;\n string s, c;\n cin >> s;\n multiset<int> ind_mst;\n for (int i = 0; i < m; ++i) {\n int a;\n cin >> a;\n ind_mst.insert(a - 1);\n }\n cin >> c;\n sort(c.begin(), c.end());\n int index = 0;\n /*\n multiset<char> char_mst;\n for (int i = 0; i < m; ++i) {\n char_mst.insert(c[i]);\n }\n */\n for (int i = 0; i < m; ++i) {\n if (ind_mst.count(*ind_mst.begin()) <= 1) {\n //char_mst.erase(prev(char_mst.end()));\n s[*ind_mst.begin()] = c[index];\n index++;\n }\n ind_mst.erase(ind_mst.begin());\n }\n cout << s << endl;\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267040279,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\n#include<iostream>\n#include<queue>\n#include<algorithm>\nusing namespace std;\nchar a[100010] = {}, b[100010] = {};\nint in[100010] = {};\t\nint n, m;\nint t;\nint main()\n{\n\tint t;\n\tcin >> t;\n\twhile (t--)\n\t{\n \n\t\tcin >> n >> m;\n\t\tcin >> a;\n\t\tfor (int i = 1; i <= m; i++)\n\t\t{\n\t\t\tcin >> in[i];\n\t\t}\n\t\tcin >> b;\n\t\tsort(in + 1, in + 1 + m);\n\t\tsort(b, b + m);\n\t\tfor (int i = 1; i <= m - 1; i++)\n\t\t{\n\t\t\tif (in[i] == in[i + 1])\n\t\t\t{\n\t\t\t\tfor (int i2 = i; i2 <= m - 1; i2++)\n\t\t\t\t{\n\t\t\t\t\tin[i2] = in[i2 + 1];\n\t\t\t\t}\n\t\t\t\tb[m - 1] = '\\0';\n\t\t\t\tm--;\n\t\t\t\ti--;\n \n\t\t\t}\n \n\t\t}\n \n\t\tfor (int i = 1; i <= m; i++)\n\t\t{\n\t\t\tif (in[i] >= 0)\n\t\t\t{\n\t\t\t\ta[in[i] - 1] = b[i - 1];\n\t\t\t}\n\t\t}\n \n\t\tputs(a);\n \n \n\t}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 269514684,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\ndef shift(c):\n c.insert(0, c.pop())\n \ndef solve():\n n, m = map(int, input().split())\n s = list(input())\n ind = sorted(map(int, input().split()))\n c = sorted(input())\n \n k = set()\n for i in range(m):\n i1 = ind[i]-1\n \n if i1 in k:\n s[i1] = min(s[i1], c[i])\n shift(c)\n else:\n s[i1] = c[i]\n \n k.add(i1)\n \n print(*s, sep=\"\")\n \nt = int(input())\nfor _ in range(t):\n solve()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267026004,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\nimport java.util.Arrays;\nimport java.util.Collection;\nimport java.util.HashSet;\nimport java.util.Scanner;\nimport java.util.TreeSet;\n \npublic class Solution2 {\n \n public static String sortString(String inputString)\n {\n char tempArray[] = inputString.toCharArray();\n \n Arrays.sort(tempArray);\n \n return new String(tempArray);\n }\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n \n int t = sc.nextInt();\n \n while(t-- > 0)\n {\n int len = sc.nextInt();\n int arr_len = sc.nextInt();\n \n String str = sc.next();\n \n int[] arr = new int[arr_len];\n \n for(int i = 0; i < arr_len; i++)\n {\n arr[i] = sc.nextInt();\n }\n \n String characters = sc.next();\n \n // put arrays into treeset\n TreeSet<Integer> treeSet = new TreeSet<>();\n \n for(int i = 0; i < arr_len; i++)\n {\n treeSet.add(arr[i]);\n }\n \n // sort the characters string in lexicographically order\n characters = sortString(characters);\n \n // replace character to lexicographically minimum character of chracters string in the str string at index of treeSet\n int k = 0;\n for(int i : treeSet)\n {\n char c = characters.charAt(k++);\n \n if(i == 1)\n {\n str = c + str.substring(1);\n }\n else if(i == len)\n {\n str = str.substring(0, len-1) + c;\n }\n else\n {\n str = str.substring(0, i-1) + c + str.substring(i);\n }\n }\n \n System.out.println(str);\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267042739,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: Update Queries\ntime_limit_ms: 2000\nmemory_limit_mb: 256\nDescription: Let's consider the following simple problem. You are given a string $s$ of length $n$, consisting of lowercase Latin letters, as well as an array of indices $ind$ of length $m$ ($1 \\leq ind_i \\leq n$) and a string $c$ of length $m$, consisting of lowercase Latin letters. Then, in order, you perform the update operations, namely, during the $i$-th operation, you set $s_{ind_i} = c_i$. Note that you perform all $m$ operations from the first to the last.\n\nOf course, if you change the order of indices in the array $ind$ and/or the order of letters in the string $c$, you can get different results. Find the lexicographically smallest string $s$ that can be obtained after $m$ update operations, if you can rearrange the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nA string $a$ is lexicographically less than a string $b$ if and only if one of the following conditions is met:\n\n * $a$ is a prefix of $b$, but $a \\neq b$; * in the first position where $a$ and $b$ differ, the symbol in string $a$ is earlier in the alphabet than the corresponding symbol in string $b$.\n\nInput: Each test consists of several sets of input data. The first line contains a single integer $t$ ($1 \\leq t \\leq 10^4$) โ the number of sets of input data. Then follows their description.\n\nThe first line of each set of input data contains two integers $n$ and $m$ ($1 \\leq n, m \\leq 10^5$) โ the length of the string $s$ and the number of updates.\n\nThe second line of each set of input data contains a string $s$ of length $n$, consisting of lowercase Latin letters.\n\nThe third line of each set of input data contains $m$ integers $ind_1, ind_2, \\ldots, ind_m$ ($1 \\leq ind_i \\leq n$) โ the array of indices $ind$.\n\nThe fourth line of each set of input data contains a string $c$ of length $m$, consisting of lowercase Latin letters.\n\nIt is guaranteed that the sum of $n$ over all sets of input data does not exceed $2 \\cdot 10^5$. Similarly, the sum of $m$ over all sets of input data does not exceed $2 \\cdot 10^5$.\n\nOutput: For each set of input data, output the lexicographically smallest string $s$ that can be obtained by rearranging the indices in the array $ind$ and the letters in the string $c$ as you like.\n\nExamples:\ninput:\n4\n1 2\na\n1 1\ncb\n4 4\nmeow\n1 2 1 4\nzcwz\n7 4\nabacaba\n1 3 5 7\ndamn\n7 10\ntraktor\n7 6 5 4 3 2 1 6 4 2\ncodeforces\noutput:\nb\ncwoz\nabdcmbn\nccdeefo\n\nNote: In the first set of input data, you can leave the array $ind$ and the string $c$ unchanged and simply perform all operations in that order.\n\nIn the second set of input data, you can set the array $ind = [1, 1, 4, 2]$ and $c =$ \"zczw\". Then the string $s$ will change as follows: $meow \\rightarrow zeow \\rightarrow ceow \\rightarrow ceoz \\rightarrow cwoz$.\n\nIn the third set of input data, you can leave the array $ind$ unchanged and set $c = $ \"admn\". Then the string $s$ will change as follows: $abacaba \\rightarrow abacaba \\rightarrow abdcaba \\rightarrow abdcmba \\rightarrow abdcmbn$.\n\nBuggy Code:\nimport java.util.*;\n \npublic class problem3 {\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt(); \n \n while (t-- > 0) {\n int n = sc.nextInt();\n int m = sc.nextInt();\n sc.nextLine();\n String s = sc.nextLine();\n int[] ind = new int[m];\n for (int i = 0; i < m; i++) {\n ind[i] = sc.nextInt();\n }\n sc.nextLine();\n String c = sc.nextLine();\n \n solve(n, m, s, ind, c);\n }\n sc.close();\n }\n \n public static void solve(int n, int m, String s, int[] ind, String c) {\n Set<Integer> st = new TreeSet<>();\n for (int i : ind) {\n st.add(i);\n }\n \n char[] cChars = c.toCharArray();\n Arrays.sort(cChars);\n String sortedT = new String(cChars);\n \n int j = 0;\n for (int i : st) {\n if (j < sortedT.length()) {\n s = s.substring(0, i - 1) + sortedT.charAt(j) + s.substring(i);\n j++;\n }\n }\n System.out.println(s);\n }\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "1986C",
"submission_id": 267266772,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\n#include <iostream>\n#include <vector>\nusing namespace std;\n \nint main() {\n int t;\n cin >> t;\n while (t--) {\n int n;\n cin>>n;\n vector<int> con(n);\n bool allZero = true, hasMiddleZero = false;\n for (int i=0; i<n; i++) {\n cin >> con[i];\n if (con[i]!=0) {\n allZero = false;\n }\n if (con[i]==0 && i>0 && i<n - 1) {\n hasMiddleZero = true;\n }\n }\n if(allZero) {\n cout<<0<<endl; \n }else if(hasMiddleZero) {\n cout<<2<<endl; \n }else{\n cout<<1<<endl; \n }\n }\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297472239,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\n#include<iostream> \n \nusing namespace std; \n \nint main(){\n // problem: https://codeforces.com/contest/2049/problem/A\n int t;\n cin >> t;\n while(t--){\n int n;\n cin >> n;\n int arr[n];\n for(int i = 0; i < n; i++){\n cin >> arr[i];\n }\n \n // if there is only one element in the array\n // if((n == 1 && arr[0] == 0) || (n == 2 && arr[0] == 0 && arr[1] == 0)){\n // cout << 0 << endl;\n // continue;\n // }\n // else if(n == 1 && arr[0] != 0) {\n // cout << 1 << endl;\n // continue;\n // }\n \n // find the number of zeros in the array\n int count = 0;\n for(int i = 0; i < n; i++){\n if(arr[i] == 0){\n count++;\n }\n }\n \n if(n == count) {\n cout << 0 << endl;\n }\n else if(count > 2) {\n cout << 2 << endl;\n }\n else if(count == 2) {\n if(arr[0] == 0 && arr[n-1] == 0)\n cout << 1 << endl;\n else \n cout << 2 << endl;\n }\n else if(count == 1) {\n if(arr[0] == 0 || arr[n-1] == 0)\n cout << 1 << endl;\n else\n cout << 2 << endl;\n }\n else {\n cout << 1 << endl;\n }\n \n }\n \n return 0 ;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 299292362,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\n#include<iostream>\n \nusing namespace std;\n \nint a[501];\n \nint main()\n{\n\tint n;\n\tcin >> n;\n\tfor (int x = 0; x < n; x++)\n\t{\n\t\tint y;//ๆต่ฏๆ ทไพ\n\t\tcin >> y;\n\t\tfor (int i = 0; i < y; i++)\n\t\t{\n\t\t\tcin >> a[i];\n\t\t}\n\t\tint temp = 0, TEMP = 0;\n\t\tfor (int m = 0; m < y; m++)\n\t\t{\n\t\t\tif (a[m] != 0) { TEMP++; }\n\t\t}\n\t\tfor (int j = 0; j < y; j++)\n\t\t{\n\t\t\tif (a[j] == 0) { temp++; }\n\t\t}\n\t\tif (TEMP == y) { cout << \"1\" << endl; continue; }\n\t\tif (temp == y) { cout << \"0\" << endl; continue; }\n \n\t\tint biao = 0;//0่กจ็คบไธญ้ดๆฒก0๏ผ1่กจ็คบไธญ้ดๆ0\t\t\n\t\tfor (int i1 = 1; i1 < y-1; i1++)\n\t\t{\n\t\t\tif (a[i1] == 0) { biao = 1; break; }\n\t\t}\n\t\t\n\t\tif (biao == 0) { cout << \"1\"<<endl; continue; }\n\t\tif (biao == 1) { cout << \"2\"<<endl; continue; }\n\t}\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297529337,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\nct = int(input())\n \nfor w in range(ct):\n l = int(input())\n \n v = list(map(int,input().split()))\n \n if sum(v) == 0:\n print(0)\n \n elif 0 not in v:\n print(1)\n \n else:\n if v[0] > 0:\n \n c = 1\n else:\n c = 0\n \n for i in range(len(v)):\n if i > 0:\n if (v[i] == 0 and v[i-1] > 0) and sum(v[i:]) > 0:\n c +=1\n \n if c > 1:\n print(2)\n else:\n print(1)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297587008,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\ndef main():\n t = int(input())\n for case in range(t):\n n = int(input())\n arr = []\n arr = list(map(int, input().split()))\n if max(arr) == 0:\n print(0)\n continue\n flag = False\n for i in range(1, len(arr)-1):\n if arr[i] == 0:\n flag2 = False\n for j in range(i+1, len(arr)):\n if arr[j]!=0:\n flag = True\n flag2 = True\n break\n if flag2:\n break\n if flag is False:\n print(1)\n else:\n print(2)\n \nmain()\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297488961,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\nimport java.util.*;\npublic class cf994a {\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n for (int i = 0; i < t; i++) {\n int n = sc.nextInt();\n int a[]= new int[n];\n int f=0,s=0;\n for (int j = 0; j < n; j++) {\n a[j] = sc.nextInt();\n s=s+a[j];\n }\n for (int j = 1; j < n-1; j++) {\n if(a[j]==0){\n f=2;\n break;\n }\n }\n if (s==0){\n f=0;\n }\n else if (f==0){\n f=1;\n }\n System.out.println(f);\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297464701,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\nimport java.util.*;\npublic class MEXDestruction {\n public static void main(String args[]){\n Scanner in = new Scanner(System.in);\n int t = in.nextInt();\n for(int i=0;i<t;i++){\n int n = in.nextInt();\n int[] arr = new int[n];\n int cnt=0;\n for(int j=0;j<n;j++){\n arr[j] = in.nextInt();\n if(arr[j]==0){\n cnt++;\n }\n }\n if(arr.length==1 && arr[0]==0){\n System.out.println(0);\n }else if(cnt==0){\n System.out.println(1);\n }\n else if(cnt==1){\n if(arr[0]==0 || arr[n-1]==0){\n System.out.println(1);\n }else{\n System.out.println(2);\n }\n }else if(cnt==n){\n System.out.println(0);\n }else if(cnt==2){\n if(arr[0]==0 && arr[n-1]==0){\n System.out.println(1);\n }else{\n System.out.println(2);\n }\n }else{\n System.out.println(2);\n }\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297626054,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\n#include<iostream>\nusing namespace std;\nint main(){\n\tint t;\n\tcin>>t;\n\twhile(t--){\n\t\tint n;\n\t\tcin>>n;\n\t\tint a[n+5];\n\t\tint fl0=0,fl1=0;\n\t\tfor(int i=1;i<=n;i++){\n\t\t\tcin>>a[i];\n\t\t\tif(a[i]==0) fl0=1;\n\t\t\telse fl1=1;\n\t\t}\n\t\tif(fl0&&!fl1){\n\t\t\tprintf(\"0\\n\");\n\t\t}\n\t\telse if(!fl0&&fl1){\n\t\t\tprintf(\"1\\n\");\n\t\t}\n\t\telse if(fl0&&fl1){\n\t\t\tint flag=0;\n\t\t\tfor(int i=1;i<=n;i++){\n\t\t\t\tif(i!=1&&i!=n&&a[i]==0&&a[i+1]!=0&&a[i-1]!=0) flag=1;\n\t\t\t}\n\t\t\tif(flag) printf(\"2\\n\");\n\t\t\telse printf(\"1\\n\");\n\t\t}\n\t}\n\t\n\treturn 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297486120,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\n#include<bits/stdc++.h>\n#include<ext/pb_ds/assoc_container.hpp>\n#include<ext/pb_ds/tree_policy.hpp>\n \nusing namespace std;\nusing namespace __gnu_pbds;\n \ntypedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds; // find_by_order, order_of_key\n#define int long long\n#define pii pair<int,int>\n#define ll long long\n#define endl '\\n'\nconst int MOD = 1e9+7;\n \n \nvoid solve()\n{\n int n; cin>>n;\n vector<int> v(n);\n int cnt=0;\n for(int i=0;i<n;i++){\n cin>>v[i];\n if(v[i]==0){\n cnt++;\n }\n }\n \n if(cnt==n){\n cout<<0<<endl;\n return;\n }else if(cnt==0){\n cout<<1<<endl;\n return;\n }\n \n for(int i=1;i<n-1;i++){\n if(v[i]==0 and v[i-1]!=0 and v[i+1]!=0){\n cout<<2<<endl;\n return;\n }\n }\n cnt=0;\n for(int i=1;i<(n-1);i++){\n if(v[i]!=0 and v[i-1]==0 and v[i+1]==0){\n cnt++;\n }\n }\n \n if(cnt>1){\n cout<<2<<endl;\n }else{\n \n cout<<1<<endl;\n }\n \n \n \n \n \n \n \n}\nint32_t main()\n{\n \n ios_base::sync_with_stdio(false);\n cin.tie(NULL);\n cout.tie(NULL);\n \n int t;\n cin>>t;\n while(t--)\n {\n solve();\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297469551,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\nusing namespace std;\n \nint main(){\n\tint t;\n\tcin >> t;\n\twhile(t--){\n\t\tint n;\n\t\tcin >> n;\n\t\tvector<int>v(n);\n\t\tint zero_cnt = 0;\n\t\tfor(int i=0 ; i<n ; i++){\n\t\t\tcin >> v[i];\n\t\t\tif(v[i] == 0) zero_cnt++;\n\t\t}\n\t\tif(zero_cnt == v.size()){\n\t\t\tcout << 0 << endl;\n\t\t\tcontinue;\n\t\t}\n\t\tif(zero_cnt == 0){\n\t\t\tcout << 1 << endl;\n\t\t\tcontinue;\n\t\t}\n\t\tint x = 0;\n\t\tint NoOfSubarray = 0;\n\t\tfor(int i=0 ; i<n ; i++){\n\t\t\tif(v[i] != 0){\n\t\t\t\tx++;\n\t\t\t}\n\t\t\tif(x == 1){\n\t\t\t\tNoOfSubarray++;\n\t\t\t}\n\t\t\tif(v[i] == 0){\n\t\t\t\tx = 0;\n\t\t\t}\n\t\t}\n\t\tif(NoOfSubarray == 1){\n\t\t\tcout << 1 << endl;\n\t\t}\n\t\telse{\n\t\t\tcout << 2 << endl;\n\t\t}\n\t}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297563508,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\ndef code(num):\n for i in range(num):\n len = int(input())\n inp = input()\n lst = [int(n) for n in inp.split(\" \")]\n count = 0\n for i in range(len):\n if lst[i] == 0:\n count += 1\n if count == len:\n print(0)\n elif count == 0:\n print(1)\n elif lst[0] == 0:\n if lst[len-1] == 0:\n if count > 2:\n print(2)\n else:\n print(1)\n elif count > 1:\n print(2)\n else:\n print(1)\n elif lst[len-1] == 0:\n if count > 1:\n print(2)\n else:\n print(1)\n else:\n print(2)\n \ncode(int(input()))\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 302188966,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\nmod = 1000000007\n \ntest_cases = int(input())\nfor _ in range(test_cases):\n size = int(input())\n arr = list(map(int, input().split()))\n zero_count = arr.count(0)\n \n if zero_count == size:\n print(0)\n else:\n if zero_count == 1 and (arr[0] == 0 or arr[-1] == 0):\n print(1)\n elif zero_count == 2 and (arr[0] == 0 and arr[-1] == 0):\n print(1)\n elif zero_count == 0:\n print(1)\n else:\n print(2)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297462931,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\nimport java.util.*;\npublic class A {\n public static void main(String[] args) {\n Scanner sc=new Scanner(System.in);\n int cases=sc.nextInt();\n while(cases-->0){\n int n=sc.nextInt();\n int[] arr=new int[n];\n int count=0;\n for(int i=0;i<n;i++){\n arr[i]=sc.nextInt();\n if(arr[i]==0){\n count++;\n }\n }\n if(count==n){\n System.out.println(0);\n }else if(count==0||check(arr)){\n System.out.println(1);\n }else{\n if(arr[0]==0&&count==1||arr[n-1]==0&&count==1||(arr[0]==0&&arr[n-1]==0&&count==2)){\n System.out.println(1);\n }else{\n System.out.println(2);\n }\n }\n \n }\n sc.close();\n \n }\n public static boolean check(int[] arr) {\n \n \n // Extract meaningful non-zero numbers for sequence validation\n int previous = -1;\n \n for (int num : arr) {\n if (num != 0) {\n if (previous != -1 && Math.abs(num - previous) != 1) {\n return false; // Sequence is broken\n }\n previous = num; // Update the last valid non-zero number\n }\n }\n \n return true;\n }\n \n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297497162,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\nimport java.util.*;\nimport java.io.*;\n \npublic class A_MEX_Destruction {\n static class FastReader {\n BufferedReader br;\n StringTokenizer st;\n \n public FastReader() {\n br = new BufferedReader(new InputStreamReader(System.in));\n }\n \n String next() {\n while (st == null || !st.hasMoreTokens()) {\n try {\n st = new StringTokenizer(br.readLine());\n } catch (IOException e) {\n e.printStackTrace();\n }\n }\n return st.nextToken();\n }\n \n int nextInt() {\n return Integer.parseInt(next());\n }\n \n long nextLong() {\n return Long.parseLong(next());\n }\n \n double nextDouble() {\n return Double.parseDouble(next());\n }\n \n String nextLine() {\n String str = \"\";\n try {\n str = br.readLine().trim();\n } catch (Exception e) {\n e.printStackTrace();\n }\n return str;\n }\n }\n \n static FastReader vt = new FastReader();\n static PrintWriter out = new PrintWriter(System.out);\n \n public static void main(String[] args) {\n try {\n int testCases = vt.nextInt();\n // int testCases = 1;\n while (testCases-- > 0) {\n // write code here\n solve();\n }\n out.flush();\n } catch (Exception e) {\n return;\n }\n }\n \n static void solve() {\n int n = vt.nextInt();\n int[] ar = new int[n];\n for (int i = 0; i < n; i++) {\n ar[i] = vt.nextInt();\n }\n int id = -1;\n int left = 0;\n int right = 0;\n for (int i = 0; i < n; i++) {\n if (ar[i] == 0 && left > 0) {\n id = i;\n break;\n } else if (ar[i] > 0) {\n left++;\n }\n }\n if (id == -1 && left != 0) {\n out.println(1);\n return;\n }\n int l = 0;\n int r = 0;\n for (int i = 0; i < id; i++) {\n if (ar[i] > 0) {\n l++;\n break;\n }\n }\n for (int i = n - 1; i > id; i--) {\n if (ar[i] > 0) {\n r++;\n } else {\n right++;\n }\n }\n if (l == 0 && r == 0) {\n out.println(0);\n } else if (l == 0 || r == 0) {\n if (right > 0) {\n out.println(2);\n return;\n }\n out.println(1);\n } else {\n out.println(2);\n }\n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297463941,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\n#include <bits/stdc++.h>\n \nusing namespace std;\n \nint main() {\n int t,n;\n cin >> t;\n while(t--) {\n cin >> n;\n vector<int> a(n);\n int sum = 0;\n for (int i = 0; i < n;i++) {\n cin >> a[i];\n sum += a[i];\n }\n if (sum == 0) {\n cout << 0 << endl;\n continue;\n }\n bool flag = 0;\n for (int i = 2; i < n;i++) {\n //cout << a[i] << \" \" << a[i-1] << \" \" << a[i-2] << endl;\n if (a[i] != 0 && a[i-1] == 0 && a[i-2] != 0) {flag = 1; \n break;\n }\n \n }\n if (flag) cout << 2 << endl;\n else cout << 1 << endl;\n }\n \n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 304624749,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\n// VK5\n#include <bits/stdc++.h>\nusing namespace std;\n// ready for the action\n#include <ext/pb_ds/assoc_container.hpp>\n#include <ext/pb_ds/tree_policy.hpp>\ntypedef __gnu_pbds::tree<int, __gnu_pbds::null_type, less<int>, __gnu_pbds::rb_tree_tag, __gnu_pbds::tree_order_statistics_node_update> ordered_set;\n#define ll long long\ntypedef pair<ll, ll> pii;\ntypedef vector<ll> vi;\ntypedef vector<ll> vll;\ntypedef vector<vi> vvi;\n#define mkp make_pair\n#define all(a) a.begin(), a.end()\n#define fsort(a) sort(a.begin(), a.end())\n#define ssort(a) sort(all(a), sortbysec);\n#define ub(a, b) upper_bound(all(a), b)\n#define lb(a, b) lower_bound(all(a), b)\n#define bs(a, b) binary_search(all(a), b)\n#define pb(a) push_back(a)\n#define sz(a) a.size()\n#define vread(a) \\\n for (auto &x : a) \\\n cin >> x\n#define vprint(a) \\\n for (auto &x : a) \\\n cout << x << ' '\n#define nl '\\n';\n#define mod 1000000007\nvoid bs_real()\n{\n float l = 0.0f, r = 1e14f;\n for (ll i = 0; i < 1000 && l + 1e-6 < r; ++i)\n {\n float mid = 0.5f * (l + r);\n // check;\n if (true)\n r = mid;\n else\n l = mid;\n }\n}\nlong long binpow(long long a, long long b)\n{ // modular exponentiation\n a %= mod;\n long long res = 1;\n while (b > 0)\n {\n if (b & 1)\n res = res * a % mod;\n a = a * a % mod;\n b >>= 1;\n }\n return res;\n}\n \nll fact(ll n)\n{\n ll ans = 1;\n for (ll i = 1; i <= n; i++)\n {\n ans = (ans * i) % mod;\n }\n return ans;\n}\nll C(ll n, ll k, ll fact[])\n{\n if (n < k)\n return 0LL;\n return (fact[n] * binpow((fact[n - k] * fact[k]) % mod, mod - 2)) % mod;\n}\n \nbool sortbysec(const pair<ll, ll> &a,\n const pair<ll, ll> &b)\n{\n if (a.second != b.second)\n return a.second < b.second;\n return a.first < b.first;\n}\n// Returns factorial of n\n \nvoid solve()\n{\n ll n;\n cin >> n;\n deque<ll> a(n), p(n);\n vread(a);\n ll h = 0;\n p = a;\n sort(all(p));\n if (p[0] == p[n - 1])\n {\n cout << 0 << endl;\n return;\n }\n \n while (a.front() == 0)\n {\n a.pop_front();\n }\n while (a.back() == 0)\n {\n a.pop_back();\n }\n ll zero = 0;\n for (ll i = 0; i < a.size(); i++)\n {\n if (a[i] == 0)\n {\n zero++;\n }\n }\n cout << 1 + (zero!=0)<< endl;\n}\n \nint32_t main()\n{\n cin.tie(nullptr);\n cout.tie(nullptr);\n ll t;\n cin >> t;\n while (t--)\n \n solve();\n return 0;\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297459667,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\nget_int = lambda: int(input())\nget_tuple_int = lambda: map(int, input().split())\nget_list_int = lambda: list(map(int, input().split()))\n \nt = get_int()\nfor _ in range(t):\n n = get_int()\n a = get_list_int()\n zeros = [i for i,j in enumerate(a) if j==0]\n if len(zeros) == n:\n print(0)\n elif len(zeros)==0:\n print(1)\n elif len(zeros)==1 and zeros[0] in [0,n-1]:\n print(1)\n elif (n-max(zeros)) + (min(zeros)+1) == len(zeros):\n print(1)\n else:\n print(2)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297461461,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\nfor _ in range(int(input())):\n t=int(input())\n w=[*map(int,input().split())]\n if w.count(0)==t:print(0)\n else:\n for i in range(t-2):\n if w[i]>0 and w[i+1]==0 and w[i+2]>0:print(2);break\n else:print(1)\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297460142,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\nimport java.util.*;\n \npublic class MEXDestruction {\n public static void main(String[] args) {\n Scanner sc = new Scanner(System.in);\n int t = sc.nextInt();\n while (t-- > 0) {\n int n = sc.nextInt();\n int[] a = new int[n];\n for (int i = 0; i < n; i++) {\n a[i] = sc.nextInt();\n }\n System.out.println(minOperations(a, n));\n }\n }\n \n private static int minOperations(int[] a, int n) {\n boolean allZero = true;\n int c=0;\n for (int num : a) {\n if (num == 0) {\n c++;\n }\n }\n if(c==0) return 1;\n if(c==n)return 0;\n boolean check=false;\n for (int i = 1; i < a.length - 1; i++) {\n if (a[i] == 0 && a[i - 1] != 0 && a[i + 1] != 0) {\n check=true;\n }\n }\n if(check)return 2;\n else return 1;\n \n }\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 297495458,
"task_type": "TestCase-Eval-Task2"
} |
TestCase-Eval-Task2 | [
{
"content": "**Task:**\nGenerate a challenging test input that exposes the bug in the buggy code of the algorithm problem:\nAlgorithm Problem:\nTitle: MEX Destruction\ntime_limit_ms: 1000\nmemory_limit_mb: 256\nDescription: Evirir the dragon snuck into a wizard's castle and found a mysterious contraption, and their playful instincts caused them to play with (destroy) it...\n\nEvirir the dragon found an array $$$a_1, a_2, \\ldots, a_n$$$ of $$$n$$$ non-negative integers.\n\nIn one operation, they can choose a non-empty subarray$$$^{\\text{โ}}$$$ $$$b$$$ of $$$a$$$ and replace it with the integer $$$\\operatorname{mex}(b)$$$$$$^{\\text{โ }}$$$. They want to use this operation any number of times to make $$$a$$$ only contain zeros. It can be proven that this is always possible under the problem constraints.\n\nWhat is the minimum number of operations needed?\n\nInput: Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \\le t \\le 200$$$). The description of the test cases follows.\n\nThe first line of each test case contains a single integer $$$n$$$ ($$$1 \\le n \\le 50$$$), the length of $$$a$$$.\n\nThe second line of each test case contains $$$n$$$ space-separated integers, $$$a_1, a_2, \\ldots, a_n$$$ ($$$0 \\le a_i \\le 100$$$).\n\nIt is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$500$$$.\n\nOutput: For each test case, output a single integer on a line, the minimum number of operations needed to make $$$a$$$ contain only zeros.\n\nExamples:\ninput:\n10\n4\n0 1 2 3\n6\n0 0 0 0 0 0\n5\n1 0 1 0 1\n5\n3 1 4 1 5\n4\n3 2 1 0\n7\n9 100 0 89 12 2 3\n4\n0 3 9 0\n7\n0 7 0 2 0 7 0\n1\n0\n2\n0 1\noutput:\n1\n0\n2\n1\n1\n2\n1\n2\n0\n1\n\nNote: In the first test case, Evirir can choose the subarray $$$b = [1, 2, 3]$$$ and replace it with $$$\\operatorname{mex}(1, 2, 3) = 0$$$, changing $$$a$$$ from $$$[0, \\underline{1, 2, 3}]$$$ to $$$[0, 0]$$$ (where the chosen subarray is underlined). Therefore, the answer is $$$1$$$.\n\nIn the second test case, $$$a$$$ already contains only $$$0$$$s, so no operation is needed.\n\nIn the third test case, Evirir can change $$$a$$$ as follows: $$$[1, \\underline{0, 1, 0, 1}] \\to [\\underline{1, 2}] \\to [0]$$$. Here, $$$\\operatorname{mex}(0, 1, 0, 1) = 2$$$ and $$$\\operatorname{mex}(1, 2) = 0$$$.\n\nIn the fourth test case, Evirir can choose $$$b$$$ to be the entire array $$$a$$$, changing $$$a$$$ from $$$[\\underline{3, 1, 4, 1, 5}]$$$ to $$$[0]$$$.\n\nBuggy Code:\nimport java.util.*;\npublic class MEX {\n public static void main(String[] args)\n {\n Scanner in=new Scanner(System.in);\n int t=in.nextInt();\n for(int z=1;z<=t;z++)\n {\n int n=in.nextInt();\n int[] a=new int[n];\n int c=0;int c1=0;int k=0;boolean flag=true;int x=0;\n for(int i=0;i<n;i++)\n {\n a[i]=in.nextInt();\n if(a[i]!=0)\n {\n x++;\n }\n if(a[i]==0 && i!=0 && i!=n-1)\n {\n c++;\n flag=false;\n }\n else if(flag==true && a[i]!=0 && i!=0 && i!=n-1)\n {\n k++;\n }\n if(a[i]==0)\n c1++;\n }\n if(c1==n)\n {\n System.out.println(\"0\");\n }\n else if(x==1 && c1==n-1)\n {\n System.out.println(\"1\");\n }\n else\n {\n if(c1-c==2||c1-c==0)\n {\n if(c>0)\n {\n System.out.println(\"2\");\n }\n else\n {\n System.out.println(\"1\");\n }\n }\n else\n {\n if(c>0 && (k+c)==n-2)\n {\n System.out.println(\"1\");\n }\n else if(c>0)\n {\n System.out.println(\"2\");\n }\n else\n {\n System.out.println(\"1\");\n }\n }\n}\n }\n}\n}\n\n\n**Instructions:**\n- Focus on edge cases or scenarios that maximize the failure probability in faulty solutions.\n- Due to the output length limit, you should generate a **small-scale** test input that is **complete and valid**.\n- Output the test input directly, not code to generate it.\n\n\n**Output format:**\n```plaintext\n{test input}\n```\n\n\nThink step by step.",
"role": "user"
}
]
| TestCase-Eval | null | {
"problem_id": "2049A",
"submission_id": 301291799,
"task_type": "TestCase-Eval-Task2"
} |
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